PHYS 231 - UNIVERSITY PHYSICS I - Circu-
lar Motion and Gravitation Question Bank - Set
2
Question 1
Question
A satellite is in a circular orbit around the Earth at a height of 500 km above
the surface. If the satellite has a mass of 1000 kg, what is the minimum velocity
required to keep the satellite in orbit?
Solution
Step 1: The minimum velocity required to keep the satellite in orbit can be deter-
mined by setting the centripetal force equal to the gravitational force acting on
the satellite. Step 2: The centripetal force acting on the satellite is provided by
the gravitational force, which can be expressed as Fcentripetal =mv2
r, where mis
the mass of the satellite, vis its velocity, and ris the distance from the center of
the Earth to the satellite’s orbit (radius of the orbit). Step 3: The gravitational
force acting on the satellite is given by Newton’s law of universal gravitation as
Fgravity =G·m·MEarth
r2, where Gis the gravitational constant, mis the mass
of the satellite, MEarth is the mass of the Earth, and ris the distance from the
center of the Earth to the satellite’s orbit. Step 4: Setting these two forces equal
to each other gives mv2
r=G·m·MEarth
r2. Step 5: Simplifying the equation, we
find v2=G·MEarth
r. Step 6: Substituting the values MEarth = 5.97 ×1024 kg,
r= 6371 km+500 km, and G= 6.67×10−11 Nm2/kg2into the equation, we can
solve for v. Step 7: Calculating rin meters as r= (6371 + 500) ×103m, we find
r= 6871 ×103m. Step 8: Substituting all values into the equation and solving
for v, we get v=r6.67 ×10−11 ×5.97 ×1024
6871 ×103. Step 9: Solving for v, we find
v≈7641 m/s. Therefore, the minimum velocity required for the satellite to stay
in orbit is approximately 7641 m/s.
Question 2
Question
A satellite moves in a circular orbit around a planet with a period of 12 hours.
The radius of the orbit is 2.0×107m. Calculate the mass of the planet.
Solution
Step 1: First, we need to find the speed of the satellite in its orbit. We can use
the formula for the speed of an object in a circular orbit:
v=2πr
T,
where vis the speed of the satellite, ris the radius of the orbit, and Tis the
period of the orbit.
Step 2: Substituting the given values, we have:
v=2π×2.0×107
12 ×3600 .
Step 3: Simplifying the expression gives:
v=8π×107
43,200 .
Step 4: Now, we find the acceleration of the satellite using the formula for
centripetal acceleration:
a=v2
r.
Step 5: Substituting the speed vfrom Step 3 and the radius rgiven in the
question, we get:
a=(8π×107
43,200 )2
2.0×107.
Step 6: Calculating the acceleration gives:
a=64π2×1014
1,856,320,000.
Step 7: Next, we can use Newton’s law of universal gravitation to find the
mass of the planet. The gravitational force between the planet and the satellite
is equal to the centripetal force required to keep the satellite in orbit:
F=mv2
r=GMm
r2,
where mis the mass of the satellite, Mis the mass of the planet, Gis the
gravitational constant, and ris the radius of the orbit.
Step 8: We can simplify the equation to solve for the mass of the planet:
M=v2r
G.
Step 9: Substituting the values for v,r, and the gravitational constant G,
we get:
M=(8π×107
43,200 )2×2.0×107
6.67 ×10−11 .
Step 10: Calculating the mass of the planet gives:
M=64π2×1014 ×2.0×107
6.67 ×10−11 .
Question 3
Question
A satellite orbits a planet in a circular path at a distance of 10,000 km from the
planet’s center. If the satellite completes one full orbit in 24 hours, determine
the mass of the planet. The gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: Find the orbital velocity of the satellite. The centripetal force required
to keep the satellite in circular motion is provided by the gravitational force
between the planet and the satellite. Equating the two forces, we have
mv2
r=GMm
r2
where - vis the orbital velocity, - ris the radius of the orbit, - Gis the grav-
itational constant, - Mis the mass of the planet, and - mis the mass of the
satellite (which cancels out).
Substitute r= 10,000 km = 10,000,000 m, G= 6.67 ×10−11 N m2/kg2, and
v=2πr
Twhere T= 24 hours = 86,400 seconds:
m2π(10,000,000)
86,400 2
10,000,000 =6.67 ×10−11M
(10,000,000)2
Solve for Min the equation above to find the mass of the planet.
Question 4
Question
A small object of mass mis attached to one end of a string that is fastened at
the other end to a nail in the ceiling. The object is given an initial velocity v0
horizontally so that it moves in a horizontal circle of radius raround the nail.
Find an expression for the tension Tin the string as a function of the object’s
speed vin its circular path. Assume the string is massless and the circular
motion is in the xy-plane.
Solution
Step 1: In circular motion, the object experiences a centripetal acceleration
towards the center of the circle given by ac=v2
r.
Step 2: The net force acting on the object towards the center of the circle is
provided by the tension in the string, which we can write as T=m·ac=m·v2
r.
Step 3: To relate the speed vto the initial velocity v0and radius r, we can
use the conservation of energy. The initial kinetic energy KE0of the object is
given by 1
2m·v2
0.
Step 4: At any point in the circular path, the total mechanical energy E
consists of kinetic energy KE =1
2m·v2and gravitational potential energy
P E = 0.
Step 5: Thus, we have KE0=KE, which gives us 1
2m·v2
0=1
2m·v2.
Step 6: From the above equation, we find v=v0.
Step 7: Substituting v=v0into the expression for tension T, we get T=
m·v2
0
r.
Therefore, the tension Tin the string as a function of the object’s speed v
in its circular path is given by T=m·v2
0
r.
Question 5
Question
A satellite is in a circular orbit around a planet. If the radius of the orbit is
doubled, how does the period of the orbit change? Justify your answer.
Solution
Step 1: Let’s denote the initial radius of the orbit as rand the period of the
orbit as T. The initial orbital velocity of the satellite can be expressed using
the centripetal force required to keep the satellite in the circular orbit:
Fcentripetal =mv2
r,
where mis the mass of the satellite, vis the velocity of the satellite, and ris
the radius.
Step 2: The centripetal force is also provided by the gravitational force
between the satellite and the planet:
Fgravity =GmM
r2,
where Mis the mass of the planet and Gis the gravitational constant.
Step 3: Equating the centripetal force and gravitational force gives:
mv2
r=GmM
r2.
Step 4: Solving for vgives us the initial orbital velocity:
v=rGM
r.
Step 5: The initial period Tof the orbit is given by:
T=2πr
v.
Step 6: Substituting the expression for vinto the equation for Tgives:
T=2πr
qGM
r
.
Step 7: If the radius of the orbit is doubled, the new radius becomes 2r.
Substituting 2rinto the equation for Tgives the new period T′:
T′=2π(2r)
qGM
(2r)
.
Step 8: Simplifying the expression for T′:
T′=4πr
qGM
(2r)
=4πr
√2qGM
r
=√2×2πr
qGM
r
=√2×T.
Step 9: Therefore, if the radius of the orbit is doubled, the period of the
orbit becomes √2 times the initial period.
Question 6
Question
A small object of mass mmoves in a horizontal circle of radius Ron a frictionless
surface. Initially, the object is moving with a speed v0. At a certain point, the
string holding the object breaks, and the object flies off tangentially. Determine
the velocity of the object just after the string breaks. Assume the string broke
at the top of the motion.
Solution
Let’s denote the velocity of the object just after the string breaks as vf. We
know that the tension in the string provides the centripetal force required for
circular motion.
Step 1: Find the centripetal acceleration just before the string breaks. The
centripetal acceleration is given by:
ac=v2
0
R
Step 2: Find the centripetal acceleration just after the string breaks. At the
top of the motion, the only force acting on the object is gravity. The object will
continue to move in a circular path with radius Rbut without any centripetal
force. The net force acting on the object is equal to the gravitational force.
m·g=m·ac
g=ac
Therefore, the centripetal acceleration just after the string breaks is equal to
the acceleration due to gravity, g.
Step 3: Find the velocity just after the string breaks. Using the equation
of motion, we can find the final velocity:
v2
f=v2
0+ 2ac·h
v2
f=v2
0+ 2g·R
v2
f=v2
0+ 2gR
Hence, the velocity of the object just after the string breaks is pv2
0+ 2gR.
Question 7
Question
A satellite of mass mis in circular orbit around a planet of mass M. The
satellite’s distance from the center of the planet is r, and the speed of the
satellite is v. Show that the gravitational force acting on the satellite is equal
to the centripetal force required for circular motion.
Solution
Step 1: The gravitational force acting on the satellite is given by Newton’s law
of universal gravitation:
Fgrav =G·M·m
r2
where Gis the gravitational constant.
Step 2: The centripetal force required for circular motion is given by the
equation:
Fcentripetal =m·v2
r
Step 3: Equate the gravitational force to the centripetal force to show that
they are equal:
G·M·m
r2=m·v2
r
Step 4: Rearrange the equation to get a more recognizable form:
G·M=v2·r2
r
Step 5: Simplify the equation:
G·M=v2·r
Therefore, the gravitational force acting on the satellite is equal to the cen-
tripetal force required for circular motion.
Question 8
Question
A satellite is in a circular orbit around Earth at an altitude of 500 km above the
surface. The satellite has a mass of 1000 kg. Calculate the speed of the satellite
in its orbit.
(Given: radius of Earth = 6.37 ×106m, mass of Earth = 5.97 ×1024 kg,
gravitational constant = 6.67 ×10−11 N m2/kg2)
Solution
Step 1: Calculate the distance from the center of the Earth to the satellite’s
orbit. The radius of the satellite’s orbit is the sum of the radius of the Earth and
the altitude of the satellite: rorbit =rEarth +h= 6.37 ×106m + 500 ×103m =
6.87 ×106m
Step 2: Calculate the gravitational force experienced by the satellite. The
gravitational force between the Earth and the satellite is given by:
F=G·mass of Earth ·mass of satellite
r2
orbit
Substitute the given values:
F=6.67 ×10−11 ·5.97 ×1024 ·1000
(6.87 ×106)2
F≈8.66 ×103N
Step 3: Calculate the centripetal force required to keep the satellite in cir-
cular motion. The centripetal force is provided by the gravitational force:
Fcentripetal =F
m·v2
rorbit
=F
1000 ·v2
6.87 ×106= 8.66 ×103
Step 4: Calculate the speed of the satellite. Solve for the speed v:
v2=8.66 ×103·6.87 ×106
1000
v≈p5.96 ×107≈7.72 ×103m/s
Thus, the speed of the satellite in its orbit is approximately 7.72 ×103m/s.
Question 9
Question
A small object of mass mis moving in a horizontal circle of radius ron a fric-
tionless surface. The object is attached to a string and is making one revolution
every Tseconds. If the string breaks when the object is at the top of its path,
find the maximum height the object will reach before falling back down.
Solution
Step 1: The force of gravity acting on the object can be split into two com-
ponents: one acting vertically downwards and one acting horizontally towards
the center of the circle. The horizontal component provides the necessary cen-
tripetal force for circular motion. The vertical component is given by mg, where
mis the mass of the object and gis the acceleration due to gravity.
Step 2: At the top of the path, the tension in the string is zero, so the net
force acting on the object in the vertical direction is only the force of gravity.
Therefore, mg provides the centripetal force required:
mv2
r=mg
Step 3: Solving for the velocity vat the top, we get:
v=√gr
Step 4: The kinetic energy of the object can be related to its potential energy
at the maximum height using the conservation of energy principle:
KEinitial +P Einitial =KEfinal +P Efinal
Since the object starts from rest at the top, KEinitial = 0.
Step 5: At the maximum height, the object has no kinetic energy, so the
total energy is equal to the potential energy:
mgh =1
2mv2
Substitute v=√gr into the equation above:
mgh =1
2m(gr)
Step 6: Solve for the maximum height h:
h=1
2r
Question 10
Question
A small object of mass mis placed on a frictionless horizontal turntable that
is rotating at a constant angular speed ω. The object remains at rest relative
to the turntable due to the force of static friction. Find the maximum possible
value of ωsuch that the object does not slide off the turntable.
Solution
Step 1: Draw a free-body diagram for the small object. There are two forces
acting on the object: the normal force (N) pointing upwards and the force of
static friction (fs) pointing towards the center of the circle.
Step 2: Write down the equations of motion in the radial direction. The net
force in the radial direction is equal to the centripetal force required to keep the
object moving in a circle. The radial acceleration aris given by ar=rα =rdω
dt ,
where ris the radius of the turntable.
Step 3: The net radial force acting on the object is given by fs=m·ar.
Since fs=µs·N, we have m·ar=µs·N, where µsis the coefficient of static
friction.
Step 4: Calculate the normal force N. Since the object is not moving verti-
cally, N=m·g, where gis the acceleration due to gravity.
Step 5: Substitute Ninto the equation m·ar=µs·Nand simplify to get
m·rdω
dt =µs·m·g.
Step 6: Rearrange the equation to solve for ω. We get dω
dt =µs·g
r.
Step 7: Integrate both sides with respect to time from 0 to ωmax to find the
maximum angular speed ωmax. We get ωmax =√2µsg.
Therefore, the maximum possible value of ωsuch that the object does not
slide off the turntable is ωmax =√2µsg.
Question 11
Question
A satellite is in a circular orbit around Earth at a height above the surface of
h= 500 km. The satellite completes one orbit in 2 hours. Calculate the orbital
speed of the satellite.
Solution
Step 1: Determine the radius of the satellite’s orbit. The altitude his given as
500 km. The radius of the orbit is the sum of the radius of Earth (REarth = 6371
km) and the altitude above the surface:
r=REarth +h= 6371 km + 500 km = 6871 km
Step 2: Calculate the mass of Earth. The gravitational force equation is
given by:
Fgrav =GmMEarth
r2
where G= 6.67×10−11 N m2/kg2is the gravitational constant. At the altitude
of 500 km, the weight of the satellite is the centripetal force that is balancing
the force of gravity:
GMEarthmsatellite
r2=msatellitev2
r
GMEarth
r2=v2
r
Solving for MEarth:
MEarth =v2r2
G
Step 3: Calculate the orbital speed of the satellite. The period Tof the orbit
is given as 2 hours.
v=2πr
T
Substitute the expression for MEarth into the equation:
MEarth =4π2r3
GT 2
v=2πr
T=2π
TrGMEarth
r
Step 4: Substitute the values into the formula. Plugging in the known values:
v=2π
2 hourss(6.67 ×10−11 N m2/kg2)(4π2(6871000 m)3)
(2 hours)2
v≈7678.2 m/s
Question 12
Question
A satellite is in a circular orbit around a planet with a radius of 5000 km. If the
satellite completes one full orbit in 8 hours, determine the mass of the planet.
Solution
Step 1: First, we need to find the speed of the satellite in its circular orbit. We
know that the speed can be calculated using the formula:
v=2πr
T
where: v= velocity of the satellite, r= radius of the orbit, and T= time taken
to complete one full orbit.
Step 2: Substituting the given values into the formula, we have:
v=2π×5000 km
8 hours
Step 3: Convert the radius to meters and the time to seconds for consistent
units:
v=2π×5000000 m
8×3600 s
Step 4: Calculate the velocity:
v=10000000π
28800 ≈1098.2 m/s
Step 5: Next, we can calculate the gravitational force acting on the satellite
using the formula:
F=mv2
r=GmM
r2
where: F= gravitational force, m= mass of the satellite, v= velocity of the
satellite, r= radius of the orbit, G= gravitational constant, and M= mass of
the planet.
Step 6: Equating the two expressions for gravitational force, we get:
mv2
r=GmM
r2
Step 7: Canceling out mand solving for M, we have:
M=v2r
G=(1098.2 m/s)2×5000000 m
6.67 ×10−11 N m2/kg2
Step 8: Calculate the mass of the planet:
M≈12036528
6.67 ×10−11 ≈1.80 ×1020 kg
Therefore, the mass of the planet is approximately 1.80 ×1020 kg.
Question 13
Question
A satellite is put into an orbit around Earth at a distance of 5000 km from
the center of Earth. The satellite completes one full orbit in 8 hours. Cal-
culate the mass of Earth given that the gravitational constant, G, is 6.674 ×
10−11 m3kg−1s−2.
Solution
Step 1: Find the orbital velocity of the satellite. The centripetal force required
to keep the satellite in circular motion is provided by the gravitational force.
The centripetal force is given by:
Fcentripetal =m×v2
r
Where: m= mass of the satellite v= orbital velocity r= distance of the satellite
from the center of Earth
The gravitational force is given by:
Fgravity =G×M×m
r2
Where: M= mass of Earth
Setting these two forces equal and solving for v:
m×v2
r=G×M×m
r2
v=rG×M
r
Given: G= 6.674 ×10−11 m3kg−1s−2r= 5000 km = 5 ×106m
Substitute these values to find v:
v=s(6.674 ×10−11 ×M)
5×106
Step 2: Calculate the orbital velocity of the satellite. Given that the satellite
completes one full orbit in 8 hours, we can find the circumference of the orbit:
Circumference = 2πr
Then, the orbital velocity is given by:
v=Circumference
Time taken for one orbit
Given: r= 5 ×106m Time taken for one orbit = 8 hours = 8×3600 seconds
Substitute these values to find v.
Step 3: Equate the two expressions for v. Equating the expressions for v
obtained from the centripetal force and the time taken for one orbit, we can
solve for M.
Step 4: Solve for the mass of Earth, M. Set the two expressions for vequal
to each other: s(6.674 ×10−11 ×M)
5×106=2π×5×106
8×3600
Solve this equation to find the mass of Earth, M.
Question 14
Question
A satellite is in circular orbit around a planet of mass M. The period of the
satellite’s orbit is Tand the radius of the orbit is r. Calculate the mass of the
planet in terms of T,r, and the universal gravitational constant G.
Solution
Step 1: Recall the formula for the acceleration of an object in circular motion
is given by: a=v2
r, where vis the velocity of the satellite.
Step 2: The velocity of the satellite can be expressed in terms of the period
Tand radius rusing the formula v=2πr
T.
Step 3: Substitute v=2πr
Tinto a=v2
rto obtain the acceleration:
a=2πr
T2
r
Step 4: Simplify the expression for acceleration to get:
a=4π2r
T2
Step 5: Recall that the gravitational force between the satellite and the
planet provides the centripetal force keeping the satellite in orbit. Therefore,
we have F=GM m
r2=ma, where mis the mass of the satellite.
Step 6: Substitute a=4π2r
T2into F=GM m
r2:
GMm
r2=m·4π2r
T2
Step 7: Cancel out the mass mon both sides and simplify the equation to
solve for the mass Mof the planet:
GM =4π2r3
T2
Step 8: Divide both sides by rto isolate M:
M=4π2r3
GT 2
Therefore, the mass of the planet in terms of T,r, and the universal gravi-
tational constant Gis 4π2r3
GT 2.
Question 15
Question
A satellite of mass mis launched into a circular orbit around the Earth at an
altitude habove the Earth’s surface. The satellite completes one orbit in a time
period T. Calculate the minimum speed the satellite must have in order to
remain in orbit.
Solution
Step 1: First, calculate the gravitational force attracting the satellite towards
the Earth. The gravitational force acting on the satellite is given by Newton’s
law of gravitation:
F=GmM
r2
where: G= 6.67×10−11 N m2/kg2is the gravitational constant, M= 5.97×1024
kg is the mass of the Earth, r=R+his the distance from the center of the
Earth to the satellite, and R= 6.37 ×106m is the radius of the Earth.
Step 2: Calculate the centripetal force acting on the satellite. The centripetal
force required to keep the satellite in circular motion is given by:
F=mv2
r
Step 3: Equate the gravitational force to the centripetal force. Setting the
gravitational force equal to the centripetal force, we have:
GmM
(R+h)2=mv2
R+h
Step 4: Solve for the minimum velocity v. Rearranging the equation from
Step 3, we get:
v=rGM
R+h
Step 5: Substitute in the given values and solve for v. Substitute the known
values into the equation in Step 4:
v=s(6.67 ×10−11 N m2/kg2)×(5.97 ×1024 kg)
(6.37 ×106m + h)
Step 6: Simplify the expression to find the minimum speed. After simplifying
the expression, the minimum velocity vrequired for the satellite to remain in
orbit is given by:
v=rGM
R+h
This is the minimum speed the satellite must have in order to remain in
orbit.
Question 16
Question
A satellite is in a circular orbit around a planet at an altitude of 10,000 km
above the planet’s surface. The satellite has a mass of 500 kg and the planet
has a mass of 5 ×1024 kg. Calculate the period of the satellite’s orbit.
(Given: The radius of the planet is 6.4×106m, and the gravitational constant
is 6.67 ×10−11 m3kg−1s−2.)
Solution
Step 1: Calculate the total radius of the satellite’s orbit including the altitude.
The total radius can be calculated by adding the altitude above the planet’s
surface to the radius of the planet:
r= 6.4×106m + 10,000 m
r= 6.41 ×106m
Step 2: Calculate the gravitational force experienced by the satellite. The
gravitational force experienced by the satellite is provided by Newton’s law of
universal gravitation:
F=G·m1·m2
r2
Substitute the given values into the equation:
F=(6.67 ×10−11 m3kg−1s−2)·(500 kg) ·(5 ×1024 kg)
(6.41 ×106m)2
F≈460 N
Step 3: Calculate the centripetal force required for circular motion. In
circular motion, the centripetal force required is provided by the gravitational
force:
Fc=m·v2
r
where mis the mass of the satellite, vis the velocity of the satellite, and ris
the radius of the orbit. Since Fc=F, we have:
m·v2
r=F
v=rF·r
m
Step 4: Calculate the velocity of the satellite. Substitute the values of F,r,
and minto the equation:
v=s460 N ·6.41 ×106m
500 kg
v≈7,134 m/s
Step 5: Calculate the period of the satellite’s orbit. The period of the
satellite’s orbit can be calculated using the formula:
T=2πr
v
Substitute the values of rand vinto the equation:
T=2π·6.41 ×106m
7,134 m/s
T≈5,672 s
T≈94.5 minutes
Therefore, the period of the satellite’s orbit is approximately 94.5 minutes.
Question 17
Question
A satellite is in a circular orbit around Earth with a radius of 10,000 km. If the
satellite completes one orbit in 12 hours, calculate the mass of Earth.
Solution
Step 1: We know that the centripetal force required to keep the satellite in
orbit is provided by the gravitational force between the satellite and Earth.
Therefore, the centripetal force is equal to the gravitational force:
mv2
r=GmM
r2
where: - mis the mass of the satellite, - vis the orbital speed of the satellite,
-ris the radius of the orbit, - Gis the gravitational constant, and - Mis the
mass of Earth.
Step 2: We need to express the orbital speed in terms of the radius of the
orbit and the period of the orbit. The orbital speed is given by:
v=2πr
T
where Tis the period of the orbit.
Step 3: Substituting the expression for velocity into the centripetal force
equation, we get:
m(2πr
T)2
r=GmM
r2
Step 4: Simplifying the equation gives us:
4π2r
T2=GM
r2
Step 5: Solving for the mass of Earth Mgives:
M=4π2r3
GT 2
Step 6: Substitute the given values of r= 10,000 km and T= 12 hours =
43,200 seconds into the equation to find M:
M=4π2×(10,000 ×103)3
6.67 ×10−11 ×(43,200)2
Step 7: Calculating the value of Mgives:
M≈5.96 ×1024 kg
Therefore, the mass of Earth is approximately 5.96 ×1024 kg.
Question 18
Question
A satellite is in a circular orbit around a planet with a radius of 2.5×107m. The
satellite completes one full orbit in 8 hours. Calculate the mass of the planet.
Solution
Step 1: Calculate the velocity of the satellite using the orbit circumference
formula.
Circumference of orbit = 2πr
Time for one orbit = 8 hours = 28800 seconds
Velocity of satellite = Circumference of orbit
Time for one orbit
Step 2: Calculate the centripetal acceleration of the satellite.
Centripetal acceleration = v2
r
Step 3: Use the centripetal acceleration formula and Newton’s law of gravi-
tation to find the mass of the planet.
Gravitational force = m×(v2/r)
r=G×m×M
r2
Where mis the mass of the satellite, Mis the mass of the planet, and Gis the
gravitational constant.
Step 4: Equate the gravitational force to the centripetal force.
G×M×m
r2=m×(v2/r)
r
Step 5: Solve for the mass of the planet.
M=v2×r
G
Question 19
Question
A satellite is in a circular orbit around Earth at an altitude of 500 km above
the surface. If the satellite has a mass of 1000 kg, what is the minimum speed
it must have to remain in orbit?
(Given: Earth’s mass = 5.97 ×1024 kg, Earth’s radius = 6371 km, Gravita-
tional constant = 6.67 ×10−11 m3/kg ·s2)
Solution
Step 1: First, we need to calculate the altitude of the satellite from the center
of the Earth. This can be done by adding Earth’s radius to the given altitude
above the surface:
Altitude = Earth’s radius+500 km = 6371 km+500 km = 6871 km = 6.871×106m
Step 2: Next, we calculate the total distance from the center of the Earth to
the satellite’s position. This is the sum of the Earth’s radius and the altitude
of the satellite:
r= Altitude + Earth’s radius = 6.871 ×106m + 6.371 ×106m = 13.242 ×106m
Step 3: Now, we can calculate the minimum speed needed for the satellite
to remain in orbit by equating the gravitational force between the Earth and
the satellite to the centripetal force required for circular motion:
GMm
r2=mv2
r
Step 4: Simplify the equation by canceling out the mass of the satellite, and
solve for the velocity v:
v=rGM
r
Step 5: Substitute the known values into the equation:
v=r6.67 ×10−11 m3/kg ·s2×5.97 ×1024 kg
13.242 ×106m
v=r39.799 ×1013
13.242 ×106
v=√3003.6
v≈54.81 m/s
Therefore, the minimum speed the satellite must have to remain in orbit is
approximately 54.81 m/s.
Question 20
Question
A satellite is in a circular orbit around the Earth at an altitude where the
acceleration due to gravity is only 2
Solution
Let’s denote the acceleration due to gravity at the Earth’s surface as gand the
acceleration due to gravity at the altitude of the satellite as 0.02g. We know
that the acceleration due to gravity is given by the formula:
g=G·MEarth
R2
Earth
,
where Gis the gravitational constant, MEarth is the mass of the Earth, and
REarth is the radius of the Earth.
Step 1: Express 0.02gin terms of gto find the altitude of the satellite.
Since the acceleration due to gravity at the altitude of the satellite is 0.02g,
we have:
0.02g=G·MEarth
(REarth +h)2,
where his the altitude of the satellite above the surface of the Earth.
Step 2: Substitute the given values into the equation and solve for h.
Substitute g=G·MEarth
R2
Earth
into the equation to get:
0.02 G·MEarth
R2
Earth =G·MEarth
(REarth +h)2.
Solving for h, we obtain:
h=q49R2
Earth −REarth = 7REarth −REarth = 6REarth = 6 ·6.37 ×106m.
Therefore, the altitude of the satellite is 38.22 ×106m.
Question 21
Question
A satellite of mass mis in a circular orbit around a planet of mass M. The
radius of the orbit is R. The satellite travels at a constant speed v. Determine
the gravitational force exerted on the satellite by the planet in terms of m,M,
R, and v.
Solution
Step 1: The centripetal force required to keep an object moving in a circular
path is provided by the gravitational force. Therefore, we have:
Fgravity =Fcentripetal
Step 2: The gravitational force between the satellite and the planet is given
by Newton’s law of universal gravitation:
Fgravity =GmM
R2
Step 3: The centripetal force required for circular motion can be expressed
as:
Fcentripetal =mv2
R
Step 4: Setting Fgravity =Fcentripetal, we have:
GmM
R2=mv2
R
Step 5: Solving for Fgravity, we get:
Fgravity =mv2
R=GmM
R2
Therefore, the gravitational force exerted on the satellite by the planet in
terms of m,M,R, and vis Fgravity =mv2
R.
Question 22
Question
A satellite is in a circular orbit around a planet with a radius of 5,000 km. If
the satellite completes one orbit in 5 hours, determine the mass of the planet.
Solution
Step 1: Find the speed of the satellite. The speed, v, of an object in circular
motion can be calculated using the formula:
v=2πr
T
where ris the radius of the orbit and Tis the time period of one complete orbit.
Substitute r= 5,000 km and T= 5 hours into the formula:
v=2π×5,000
5
v=10,000π
5
v= 2,000πkm/h
Step 2: Use the gravitational force formula. The centripetal force required
to keep the satellite in orbit is provided by the gravitational force between the
satellite and the planet. This can be calculated using the formula:
F=Gm1m2
r2
where Fis the gravitational force, Gis the gravitational constant, m1is the
mass of the planet, m2is the mass of the satellite, and ris the distance between
the two objects. The gravitational force is also equal to the centripetal force:
F=m2v2
r
Equating the two expressions for Fgives:
Gm1m2
r2=m2v2
r
Gm1=v2r
We are interested in finding the mass of the planet, so we will solve for m1:
m1=v2r
G
Substitute v= 2,000πkm/h, r= 5,000 km, and G= 6.67 ×10−11 m3/kg/s2
into the formula:
m1=(2,000π)2×5,000
6.67 ×10−11
m1=4,000,000π2×5,000
6.67 ×10−11
m1=20,000,000π2
6.67 ×10−11 ≈2.39 ×1021 kg
Therefore, the mass of the planet is approximately 2.39 ×1021 kg.
Question 23
Question
A satellite orbits a planet in a circular path with a radius of 2.5×107meters.
If the satellite completes one orbit in 30 hours, what is the acceleration of the
satellite towards the planet? Assume the orbit is close to the surface of the
planet.
(Given: Gravitational constant G= 6.67 ×10−11 m3/kg ·s2, mass of the
planet M= 5.97 ×1024 kg, radius of the planet r= 6.37 ×106meters)
Solution
Step 1: First, we need to find the speed of the satellite. We know that the time
period of one orbit is 30 hours, so the angular velocity, ω, can be calculated as:
ω=2π
T=2π
30 ×60 ×60 ≈5.79 ×10−5s−1
Step 2: The velocity of the satellite in circular motion can be found using
the formula v=rω. Substituting the values, we get:
v= 2.5×107×5.79 ×10−5≈1447.5 m/s
Step 3: Next, we can calculate the centripetal acceleration of the satellite
using the formula a=v2
r. Substituting the values, we get:
a=(1447.5)2
2.5×107≈1.029 m/s2
Therefore, the acceleration of the satellite towards the planet is approxi-
mately 1.029 m/s2.
Question 24
Question
A satellite is in a circular orbit around Earth with a radius of 8000 km. De-
termine the speed of the satellite in its orbit. Given that the mass of Earth is
5.97 ×1024 kg and the gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: The centripetal force required for the satellite to stay in its circular
orbit is provided by the gravitational force between the satellite and Earth.
Therefore, we have:
m·v2
r=G·M·m
r2
where: m= mass of the satellite v= speed of the satellite r= radius of the
satellite’s orbit M= mass of Earth G= gravitational constant
Step 2: We can cancel out the mass of the satellite, m, from both sides of
the equation:
v2
r=G·M
r2
Step 3: Rearranging the equation to solve for the speed, v, we get:
v=rG·M
r
Step 4: Substituting the known values into the equation, we find:
v=s(6.67 ×10−11 N m2/kg2)·(5.97 ×1024 kg)
8000 ×103m
Step 5: Calculating the speed, we get:
v=r4.00239 ×1014
8×106=p50.029875 ×107= 2236.067998 m/s
Therefore, the speed of the satellite in its circular orbit is approximately
2236.07 m/s.
Question 25
Question
A small object of mass mis attached to a string of length Land is swung in a
horizontal circle with a constant speed v. If the string makes an angle θwith
the vertical direction, find the tension in the string in terms of m,L,v, and θ.
Assume no air resistance.
Solution
Step 1: Draw a free-body diagram of the object in circular motion. The forces
acting on the object are the tension Tin the string, the gravitational force
mg pointing downward, and the centripetal force mv2/L pointing towards the
center of the circle.
Step 2: Resolve the forces into their component directions. The gravita-
tional force mg has a vertical component mg cos(θ) and a horizontal component
mg sin(θ). The tension Thas a horizontal component Tcos(θ) and a vertical
component Tsin(θ).
Step 3: Write the force balance equations in the vertical and horizontal direc-
tions: Vertical direction: Tsin(θ) = mg cos(θ) Horizontal direction: Tcos(θ) =
mv2
L
Step 4: Solve the equation Tsin(θ) = mg cos(θ) for T:T=mg cos(θ)
sin(θ)
Step 5: Substitute the expression for Tinto the horizontal force balance
equation: mg cos(θ)
sin(θ)cos(θ) = mv2
L
Step 6: Simplify the equation to solve for T:T=mv2
Lsin(θ)
Therefore, the tension in the string is mv2
Lsin(θ).
Question 26
Question
A satellite is in a circular orbit around Earth at an altitude of 500 km above
the surface. If the satellite’s speed is 7.5 km/s, what is its period of revolution?
Given that the radius of Earth is 6370 km and its mass is 5.97 ×1024 kg.
Solution
Step 1: Calculate the total distance the satellite travels in one revolution.
The total distance can be calculated as the circumference of the circular
orbit. The radius of the orbit is the sum of the radius of Earth and the altitude
of the satellite:
R= 6370 km + 500 km = 6870 km
The circumference is given by 2πR.
Step 2: Calculate the period of revolution using the distance traveled and
the satellite’s speed.
The speed of the satellite is equal to the distance traveled divided by the
time taken:
v=2πR
T
Where: - v= 7.5 km/s (speed of the satellite), - R= 6870 km (radius of the
orbit).
Rearranging the formula to solve for the period of revolution T:
T=2πR
v
Substitute the given values:
T=2π×6870
7.5s
T≈2×3.14159 ×6870
7.5≈3279.37 s
Hence, the period of revolution for the satellite is approximately 3279.37 sec-
onds.
Question 27
Question
A satellite is in a circular orbit around Earth at an altitude of 1000 km above
the surface. If the radius of Earth is 6400 km, calculate the satellite’s orbital
speed. Assume the satellite is very far away from the effects of the Earth’s
atmosphere.
Solution
Step 1: We can start by finding the total distance from the center of the Earth to
the satellite’s orbit. This is the sum of the radius of the Earth and the altitude
of the satellite above the surface.
r= 6400 km + 1000 km = 7400 km
Step 2: We can now apply the gravitational force equation to find the satel-
lite’s orbital speed. The gravitational force is equal to the centripetal force
required to keep the satellite in its orbit.
GMm
74002=mv2
7400
Where: - G= 6.67 ×10−11 N m2/kg2(Gravitational constant) - M= 5.97 ×
1024 kg (Mass of Earth) - m= mass of the satellite (which cancels out) - v=
orbital speed
Step 3: We can simplify the equation and solve for v.
v=rGM
r
v=s(6.67 ×10−11 N m2/kg2)(5.97 ×1024 kg)
7400 km ×103m/km
v=r39.8×1013
7400 ×103
v≈p5.37 ×106m/s
v≈2317 m/s
Therefore, the satellite’s orbital speed is approximately 2317 m/s.
Question 28
Question
A satellite is in an elliptical orbit around the Earth. At the point where it is
closest to the Earth (perigee), the speed of the satellite is 4.5×103m/s. At the
point where it is farthest from the Earth (apogee), the speed is 2.5×103m/s.
Calculate the total mechanical energy of the satellite.
Given: Mass of the Earth, M= 5.97 ×1024 kg
Radius of the Earth, R= 6.37 ×106m
Solution
Step 1: Find the semi-major axis of the orbit using the information provided.
At perigee (closest to the Earth): v1= 4.5×103m/s
At apogee (farthest from the Earth): v2= 2.5×103m/s
Using conservation of angular momentum, we have:
r1·v1=r2·v2
where r1and r2are the distances at perigee and apogee respectively.
Step 2: Find the velocities v1and v2at perigee and apogee. The kinetic
energy of the satellite is given by:
KE =1
2mv2
At perigee:
KE1=1
2mv2
1
At apogee:
KE2=1
2mv2
2
Step 3: Find the potential energy of the satellite at each point. The potential
energy of the satellite at a distance rfrom the center of the Earth is given by:
P E =−GMm
r
where Mis the mass of the Earth, Gis the gravitational constant, and mis the
mass of the satellite.
Step 4: Calculate the total mechanical energy of the satellite. The total
mechanical energy Eof the satellite is the sum of its kinetic and potential
energies:
E=KE +P E
Question 29
Question
A satellite is in a circular orbit around the Earth at a distance of 8000 km from
the center of the Earth. If the satellite completes one orbit in 10 hours, calculate
the mass of the Earth. Given G= 6.67 ×10−11 m3kg−1s−2.
Solution
Step 1: First, calculate the velocity of the satellite in its orbit using the formula
for the speed of an object in circular motion. The centripetal force required to
keep the satellite in orbit is provided by the gravitational force.
v=2πr
T
where: v= velocity of the satellite, r= radius of the orbit, T= time period of
one orbit.
Given r= 8000 km and T= 10 hours, convert rto meters and Tto seconds
before substituting into the formula.
Step 2: Calculate the centripetal acceleration of the satellite.
ac=v2
r
Step 3: Equate the centripetal acceleration to the gravitational acceleration,
and solve for the mass of the Earth using Newton’s law of gravitation.
Fc=Fg
mv2
r=GmM
r2
where: Fc= centripetal force, Fg= gravitational force, m= mass of the satellite,
M= mass of the Earth.
Step 4: Substitute the values of vand rinto the equation and solve for M,
the mass of the Earth.
Question 30
Question
A satellite in a circular orbit around Earth has a period of revolution of 12 hours.
If the radius of the orbit is 3 ×107m, find the mass of the Earth. Assume the
satellite is in a vacuum and G= 6.67 ×10−11 m3kg−1s−2.
Solution
Step 1: First, let’s calculate the speed of the satellite using the formula for the
centripetal force in circular motion:
Fcentripetal =m·v2
r
m·v2
r=G·M·m
r2
v2=G·M
r
v=rG·M
r
where: - Fcentripetal is the centripetal force, - mis the mass of the satellite, -
vis the speed of the satellite, - ris the radius of the satellite’s orbit, - Gis the
gravitational constant, - Mis the mass of the Earth.
Step 2: Now, let’s find the speed of the satellite by plugging in the known
values:
v=s(6.67 ×10−11 m3kg−1s−2)·M
3×107
v=r2.223 ×103·M
107
Step 3: Since the period of revolution is given by T=2πr
v, we can express v
in terms of T:
v=2πr
T
Step 4: Substitute the expression for vback into the previous equation:
r2.223 ×103·M
107=2π·3×107
12 ×60 ×60
2.223 ×103·M
107=2π·3×107
12 ×60 ×602
Question 3
Question
A satellite orbits a planet in a circular path at a distance of 10,000 km from the
planet’s center. If the satellite completes one full orbit in 24 hours, determine
the mass of the planet. The gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: Find the orbital velocity of the satellite. The centripetal force required
to keep the satellite in circular motion is provided by the gravitational force
between the planet and the satellite. Equating the two forces, we have
mv2
r=GMm
r2
where - vis the orbital velocity, - ris the radius of the orbit, - Gis the grav-
itational constant, - Mis the mass of the planet, and - mis the mass of the
satellite (which cancels out).
Substitute r= 10,000 km = 10,000,000 m, G= 6.67 ×10−11 N m2/kg2, and
v=2πr
Twhere T= 24 hours = 86,400 seconds:
m2π(10,000,000)
86,400 2
10,000,000 =6.67 ×10−11M
(10,000,000)2
Solve for Min the equation above to find the mass of the planet.
Question 4
Question
A small object of mass mis attached to one end of a string that is fastened at
the other end to a nail in the ceiling. The object is given an initial velocity v0
horizontally so that it moves in a horizontal circle of radius raround the nail.
Find an expression for the tension Tin the string as a function of the object’s
speed vin its circular path. Assume the string is massless and the circular
motion is in the xy-plane.
Solution
Step 1: In circular motion, the object experiences a centripetal acceleration
towards the center of the circle given by ac=v2
r.
Step 2: The net force acting on the object towards the center of the circle is
provided by the tension in the string, which we can write as T=m·ac=m·v2
r.
Step 3: To relate the speed vto the initial velocity v0and radius r, we can
use the conservation of energy. The initial kinetic energy KE0of the object is
given by 1
2m·v2
0.
Step 4: At any point in the circular path, the total mechanical energy E
consists of kinetic energy KE =1
2m·v2and gravitational potential energy
P E = 0.
Step 5: Thus, we have KE0=KE, which gives us 1
2m·v2
0=1
2m·v2.
Step 6: From the above equation, we find v=v0.
Step 7: Substituting v=v0into the expression for tension T, we get T=
m·v2
0
r.
Therefore, the tension Tin the string as a function of the object’s speed v
in its circular path is given by T=m·v2
0
r.
Question 5
Question
A satellite is in a circular orbit around a planet. If the radius of the orbit is
doubled, how does the period of the orbit change? Justify your answer.
Solution
Step 1: Let’s denote the initial radius of the orbit as rand the period of the
orbit as T. The initial orbital velocity of the satellite can be expressed using
the centripetal force required to keep the satellite in the circular orbit:
Fcentripetal =mv2
r,
where mis the mass of the satellite, vis the velocity of the satellite, and ris
the radius.
Step 2: The centripetal force is also provided by the gravitational force
between the satellite and the planet:
Fgravity =GmM
r2,
where Mis the mass of the planet and Gis the gravitational constant.
Step 3: Equating the centripetal force and gravitational force gives:
mv2
r=GmM
r2.
Step 4: Solving for vgives us the initial orbital velocity:
v=rGM
r.
Step 5: The initial period Tof the orbit is given by:
T=2πr
v.
Step 6: Substituting the expression for vinto the equation for Tgives:
T=2πr
qGM
r
.
Step 7: If the radius of the orbit is doubled, the new radius becomes 2r.
Substituting 2rinto the equation for Tgives the new period T′:
T′=2π(2r)
qGM
(2r)
.
Step 8: Simplifying the expression for T′:
T′=4πr
qGM
(2r)
=4πr
√2qGM
r
=√2×2πr
qGM
r
=√2×T.
Step 9: Therefore, if the radius of the orbit is doubled, the period of the
orbit becomes √2 times the initial period.
Question 6
Question
A small object of mass mmoves in a horizontal circle of radius Ron a frictionless
surface. Initially, the object is moving with a speed v0. At a certain point, the
string holding the object breaks, and the object flies off tangentially. Determine
the velocity of the object just after the string breaks. Assume the string broke
at the top of the motion.
Solution
Let’s denote the velocity of the object just after the string breaks as vf. We
know that the tension in the string provides the centripetal force required for
circular motion.
Step 1: Find the centripetal acceleration just before the string breaks. The
centripetal acceleration is given by:
ac=v2
0
R
Step 2: Find the centripetal acceleration just after the string breaks. At the
top of the motion, the only force acting on the object is gravity. The object will
continue to move in a circular path with radius Rbut without any centripetal
force. The net force acting on the object is equal to the gravitational force.
m·g=m·ac
g=ac
Therefore, the centripetal acceleration just after the string breaks is equal to
the acceleration due to gravity, g.
Step 3: Find the velocity just after the string breaks. Using the equation
of motion, we can find the final velocity:
v2
f=v2
0+ 2ac·h
v2
f=v2
0+ 2g·R
v2
f=v2
0+ 2gR
Hence, the velocity of the object just after the string breaks is pv2
0+ 2gR.
Question 7
Question
A satellite of mass mis in circular orbit around a planet of mass M. The
satellite’s distance from the center of the planet is r, and the speed of the
satellite is v. Show that the gravitational force acting on the satellite is equal
to the centripetal force required for circular motion.
Solution
Step 1: The gravitational force acting on the satellite is given by Newton’s law
of universal gravitation:
Fgrav =G·M·m
r2
where Gis the gravitational constant.
Step 2: The centripetal force required for circular motion is given by the
equation:
Fcentripetal =m·v2
r
Step 3: Equate the gravitational force to the centripetal force to show that
they are equal:
G·M·m
r2=m·v2
r
Step 4: Rearrange the equation to get a more recognizable form:
G·M=v2·r2
r
Step 5: Simplify the equation:
G·M=v2·r
Therefore, the gravitational force acting on the satellite is equal to the cen-
tripetal force required for circular motion.
Question 8
Question
A satellite is in a circular orbit around Earth at an altitude of 500 km above the
surface. The satellite has a mass of 1000 kg. Calculate the speed of the satellite
in its orbit.
(Given: radius of Earth = 6.37 ×106m, mass of Earth = 5.97 ×1024 kg,
gravitational constant = 6.67 ×10−11 N m2/kg2)
Solution
Step 1: Calculate the distance from the center of the Earth to the satellite’s
orbit. The radius of the satellite’s orbit is the sum of the radius of the Earth and
the altitude of the satellite: rorbit =rEarth +h= 6.37 ×106m + 500 ×103m =
6.87 ×106m
Step 2: Calculate the gravitational force experienced by the satellite. The
gravitational force between the Earth and the satellite is given by:
F=G·mass of Earth ·mass of satellite
r2
orbit
Substitute the given values:
F=6.67 ×10−11 ·5.97 ×1024 ·1000
(6.87 ×106)2
F≈8.66 ×103N
Step 3: Calculate the centripetal force required to keep the satellite in cir-
cular motion. The centripetal force is provided by the gravitational force:
Fcentripetal =F
m·v2
rorbit
=F
1000 ·v2
6.87 ×106= 8.66 ×103
Step 4: Calculate the speed of the satellite. Solve for the speed v:
v2=8.66 ×103·6.87 ×106
1000
v≈p5.96 ×107≈7.72 ×103m/s
Thus, the speed of the satellite in its orbit is approximately 7.72 ×103m/s.
Question 9
Question
A small object of mass mis moving in a horizontal circle of radius ron a fric-
tionless surface. The object is attached to a string and is making one revolution
every Tseconds. If the string breaks when the object is at the top of its path,
find the maximum height the object will reach before falling back down.
Solution
Step 1: The force of gravity acting on the object can be split into two com-
ponents: one acting vertically downwards and one acting horizontally towards
the center of the circle. The horizontal component provides the necessary cen-
tripetal force for circular motion. The vertical component is given by mg, where
mis the mass of the object and gis the acceleration due to gravity.
Step 2: At the top of the path, the tension in the string is zero, so the net
force acting on the object in the vertical direction is only the force of gravity.
Therefore, mg provides the centripetal force required:
mv2
r=mg
Step 3: Solving for the velocity vat the top, we get:
v=√gr
Step 4: The kinetic energy of the object can be related to its potential energy
at the maximum height using the conservation of energy principle:
KEinitial +P Einitial =KEfinal +P Efinal
Since the object starts from rest at the top, KEinitial = 0.
Step 5: At the maximum height, the object has no kinetic energy, so the
total energy is equal to the potential energy:
mgh =1
2mv2
Substitute v=√gr into the equation above:
mgh =1
2m(gr)
Step 6: Solve for the maximum height h:
h=1
2r
Question 10
Question
A small object of mass mis placed on a frictionless horizontal turntable that
is rotating at a constant angular speed ω. The object remains at rest relative
to the turntable due to the force of static friction. Find the maximum possible
value of ωsuch that the object does not slide off the turntable.
Solution
Step 1: Draw a free-body diagram for the small object. There are two forces
acting on the object: the normal force (N) pointing upwards and the force of
static friction (fs) pointing towards the center of the circle.
Step 2: Write down the equations of motion in the radial direction. The net
force in the radial direction is equal to the centripetal force required to keep the
object moving in a circle. The radial acceleration aris given by ar=rα =rdω
dt ,
where ris the radius of the turntable.
Step 3: The net radial force acting on the object is given by fs=m·ar.
Since fs=µs·N, we have m·ar=µs·N, where µsis the coefficient of static
friction.
Step 4: Calculate the normal force N. Since the object is not moving verti-
cally, N=m·g, where gis the acceleration due to gravity.
Step 5: Substitute Ninto the equation m·ar=µs·Nand simplify to get
m·rdω
dt =µs·m·g.
Step 6: Rearrange the equation to solve for ω. We get dω
dt =µs·g
r.
Step 7: Integrate both sides with respect to time from 0 to ωmax to find the
maximum angular speed ωmax. We get ωmax =√2µsg.
Therefore, the maximum possible value of ωsuch that the object does not
slide off the turntable is ωmax =√2µsg.
Question 11
Question
A satellite is in a circular orbit around Earth at a height above the surface of
h= 500 km. The satellite completes one orbit in 2 hours. Calculate the orbital
speed of the satellite.
Solution
Step 1: Determine the radius of the satellite’s orbit. The altitude his given as
500 km. The radius of the orbit is the sum of the radius of Earth (REarth = 6371
km) and the altitude above the surface:
r=REarth +h= 6371 km + 500 km = 6871 km
Step 2: Calculate the mass of Earth. The gravitational force equation is
given by:
Fgrav =GmMEarth
r2
where G= 6.67×10−11 N m2/kg2is the gravitational constant. At the altitude
of 500 km, the weight of the satellite is the centripetal force that is balancing
the force of gravity:
GMEarthmsatellite
r2=msatellitev2
r
GMEarth
r2=v2
r
Solving for MEarth:
MEarth =v2r2
G
Step 3: Calculate the orbital speed of the satellite. The period Tof the orbit
is given as 2 hours.
v=2πr
T
Substitute the expression for MEarth into the equation:
MEarth =4π2r3
GT 2
v=2πr
T=2π
TrGMEarth
r
Step 4: Substitute the values into the formula. Plugging in the known values:
v=2π
2 hourss(6.67 ×10−11 N m2/kg2)(4π2(6871000 m)3)
(2 hours)2
v≈7678.2 m/s
Question 12
Question
A satellite is in a circular orbit around a planet with a radius of 5000 km. If the
satellite completes one full orbit in 8 hours, determine the mass of the planet.
Solution
Step 1: First, we need to find the speed of the satellite in its circular orbit. We
know that the speed can be calculated using the formula:
v=2πr
T
where: v= velocity of the satellite, r= radius of the orbit, and T= time taken
to complete one full orbit.
Step 2: Substituting the given values into the formula, we have:
v=2π×5000 km
8 hours
Step 3: Convert the radius to meters and the time to seconds for consistent
units:
v=2π×5000000 m
8×3600 s
Step 4: Calculate the velocity:
v=10000000π
28800 ≈1098.2 m/s
Step 5: Next, we can calculate the gravitational force acting on the satellite
using the formula:
F=mv2
r=GmM
r2
where: F= gravitational force, m= mass of the satellite, v= velocity of the
satellite, r= radius of the orbit, G= gravitational constant, and M= mass of
the planet.
Step 6: Equating the two expressions for gravitational force, we get:
mv2
r=GmM
r2
Step 7: Canceling out mand solving for M, we have:
M=v2r
G=(1098.2 m/s)2×5000000 m
6.67 ×10−11 N m2/kg2
Step 8: Calculate the mass of the planet:
M≈12036528
6.67 ×10−11 ≈1.80 ×1020 kg
Therefore, the mass of the planet is approximately 1.80 ×1020 kg.
Question 13
Question
A satellite is put into an orbit around Earth at a distance of 5000 km from
the center of Earth. The satellite completes one full orbit in 8 hours. Cal-
culate the mass of Earth given that the gravitational constant, G, is 6.674 ×
10−11 m3kg−1s−2.
Solution
Step 1: Find the orbital velocity of the satellite. The centripetal force required
to keep the satellite in circular motion is provided by the gravitational force.
The centripetal force is given by:
Fcentripetal =m×v2
r
Where: m= mass of the satellite v= orbital velocity r= distance of the satellite
from the center of Earth
The gravitational force is given by:
Fgravity =G×M×m
r2
Where: M= mass of Earth
Setting these two forces equal and solving for v:
m×v2
r=G×M×m
r2
v=rG×M
r
Given: G= 6.674 ×10−11 m3kg−1s−2r= 5000 km = 5 ×106m
Substitute these values to find v:
v=s(6.674 ×10−11 ×M)
5×106
Step 2: Calculate the orbital velocity of the satellite. Given that the satellite
completes one full orbit in 8 hours, we can find the circumference of the orbit:
Circumference = 2πr
Then, the orbital velocity is given by:
v=Circumference
Time taken for one orbit
Given: r= 5 ×106m Time taken for one orbit = 8 hours = 8×3600 seconds
Substitute these values to find v.
Step 3: Equate the two expressions for v. Equating the expressions for v
obtained from the centripetal force and the time taken for one orbit, we can
solve for M.
Step 4: Solve for the mass of Earth, M. Set the two expressions for vequal
to each other: s(6.674 ×10−11 ×M)
5×106=2π×5×106
8×3600
Solve this equation to find the mass of Earth, M.
Question 14
Question
A satellite is in circular orbit around a planet of mass M. The period of the
satellite’s orbit is Tand the radius of the orbit is r. Calculate the mass of the
planet in terms of T,r, and the universal gravitational constant G.
Solution
Step 1: Recall the formula for the acceleration of an object in circular motion
is given by: a=v2
r, where vis the velocity of the satellite.
Step 2: The velocity of the satellite can be expressed in terms of the period
Tand radius rusing the formula v=2πr
T.
Step 3: Substitute v=2πr
Tinto a=v2
rto obtain the acceleration:
a=2πr
T2
r
Step 4: Simplify the expression for acceleration to get:
a=4π2r
T2
Step 5: Recall that the gravitational force between the satellite and the
planet provides the centripetal force keeping the satellite in orbit. Therefore,
we have F=GM m
r2=ma, where mis the mass of the satellite.
Step 6: Substitute a=4π2r
T2into F=GM m
r2:
GMm
r2=m·4π2r
T2
Step 7: Cancel out the mass mon both sides and simplify the equation to
solve for the mass Mof the planet:
GM =4π2r3
T2
Step 8: Divide both sides by rto isolate M:
M=4π2r3
GT 2
Therefore, the mass of the planet in terms of T,r, and the universal gravi-
tational constant Gis 4π2r3
GT 2.
Question 15
Question
A satellite of mass mis launched into a circular orbit around the Earth at an
altitude habove the Earth’s surface. The satellite completes one orbit in a time
period T. Calculate the minimum speed the satellite must have in order to
remain in orbit.
Solution
Step 1: First, calculate the gravitational force attracting the satellite towards
the Earth. The gravitational force acting on the satellite is given by Newton’s
law of gravitation:
F=GmM
r2
where: G= 6.67×10−11 N m2/kg2is the gravitational constant, M= 5.97×1024
kg is the mass of the Earth, r=R+his the distance from the center of the
Earth to the satellite, and R= 6.37 ×106m is the radius of the Earth.
Step 2: Calculate the centripetal force acting on the satellite. The centripetal
force required to keep the satellite in circular motion is given by:
F=mv2
r
Step 3: Equate the gravitational force to the centripetal force. Setting the
gravitational force equal to the centripetal force, we have:
GmM
(R+h)2=mv2
R+h
Step 4: Solve for the minimum velocity v. Rearranging the equation from
Step 3, we get:
v=rGM
R+h
Step 5: Substitute in the given values and solve for v. Substitute the known
values into the equation in Step 4:
v=s(6.67 ×10−11 N m2/kg2)×(5.97 ×1024 kg)
(6.37 ×106m + h)
Step 6: Simplify the expression to find the minimum speed. After simplifying
the expression, the minimum velocity vrequired for the satellite to remain in
orbit is given by:
v=rGM
R+h
This is the minimum speed the satellite must have in order to remain in
orbit.
Question 16
Question
A satellite is in a circular orbit around a planet at an altitude of 10,000 km
above the planet’s surface. The satellite has a mass of 500 kg and the planet
has a mass of 5 ×1024 kg. Calculate the period of the satellite’s orbit.
(Given: The radius of the planet is 6.4×106m, and the gravitational constant
is 6.67 ×10−11 m3kg−1s−2.)
Solution
Step 1: Calculate the total radius of the satellite’s orbit including the altitude.
The total radius can be calculated by adding the altitude above the planet’s
surface to the radius of the planet:
r= 6.4×106m + 10,000 m
r= 6.41 ×106m
Step 2: Calculate the gravitational force experienced by the satellite. The
gravitational force experienced by the satellite is provided by Newton’s law of
universal gravitation:
F=G·m1·m2
r2
Substitute the given values into the equation:
F=(6.67 ×10−11 m3kg−1s−2)·(500 kg) ·(5 ×1024 kg)
(6.41 ×106m)2
F≈460 N
Step 3: Calculate the centripetal force required for circular motion. In
circular motion, the centripetal force required is provided by the gravitational
force:
Fc=m·v2
r
where mis the mass of the satellite, vis the velocity of the satellite, and ris
the radius of the orbit. Since Fc=F, we have:
m·v2
r=F
v=rF·r
m
Step 4: Calculate the velocity of the satellite. Substitute the values of F,r,
and minto the equation:
v=s460 N ·6.41 ×106m
500 kg
v≈7,134 m/s
Step 5: Calculate the period of the satellite’s orbit. The period of the
satellite’s orbit can be calculated using the formula:
T=2πr
v
Substitute the values of rand vinto the equation:
T=2π·6.41 ×106m
7,134 m/s
T≈5,672 s
T≈94.5 minutes
Therefore, the period of the satellite’s orbit is approximately 94.5 minutes.
Question 17
Question
A satellite is in a circular orbit around Earth with a radius of 10,000 km. If the
satellite completes one orbit in 12 hours, calculate the mass of Earth.
Solution
Step 1: We know that the centripetal force required to keep the satellite in
orbit is provided by the gravitational force between the satellite and Earth.
Therefore, the centripetal force is equal to the gravitational force:
mv2
r=GmM
r2
where: - mis the mass of the satellite, - vis the orbital speed of the satellite,
-ris the radius of the orbit, - Gis the gravitational constant, and - Mis the
mass of Earth.
Step 2: We need to express the orbital speed in terms of the radius of the
orbit and the period of the orbit. The orbital speed is given by:
v=2πr
T
where Tis the period of the orbit.
Step 3: Substituting the expression for velocity into the centripetal force
equation, we get:
m(2πr
T)2
r=GmM
r2
Step 4: Simplifying the equation gives us:
4π2r
T2=GM
r2
Step 5: Solving for the mass of Earth Mgives:
M=4π2r3
GT 2
Step 6: Substitute the given values of r= 10,000 km and T= 12 hours =
43,200 seconds into the equation to find M:
M=4π2×(10,000 ×103)3
6.67 ×10−11 ×(43,200)2
Step 7: Calculating the value of Mgives:
M≈5.96 ×1024 kg
Therefore, the mass of Earth is approximately 5.96 ×1024 kg.
Question 18
Question
A satellite is in a circular orbit around a planet with a radius of 2.5×107m. The
satellite completes one full orbit in 8 hours. Calculate the mass of the planet.
Solution
Step 1: Calculate the velocity of the satellite using the orbit circumference
formula.
Circumference of orbit = 2πr
Time for one orbit = 8 hours = 28800 seconds
Velocity of satellite = Circumference of orbit
Time for one orbit
Step 2: Calculate the centripetal acceleration of the satellite.
Centripetal acceleration = v2
r
Step 3: Use the centripetal acceleration formula and Newton’s law of gravi-
tation to find the mass of the planet.
Gravitational force = m×(v2/r)
r=G×m×M
r2
Where mis the mass of the satellite, Mis the mass of the planet, and Gis the
gravitational constant.
Step 4: Equate the gravitational force to the centripetal force.
G×M×m
r2=m×(v2/r)
r
Step 5: Solve for the mass of the planet.
M=v2×r
G
Question 19
Question
A satellite is in a circular orbit around Earth at an altitude of 500 km above
the surface. If the satellite has a mass of 1000 kg, what is the minimum speed
it must have to remain in orbit?
(Given: Earth’s mass = 5.97 ×1024 kg, Earth’s radius = 6371 km, Gravita-
tional constant = 6.67 ×10−11 m3/kg ·s2)
Solution
Step 1: First, we need to calculate the altitude of the satellite from the center
of the Earth. This can be done by adding Earth’s radius to the given altitude
above the surface:
Altitude = Earth’s radius+500 km = 6371 km+500 km = 6871 km = 6.871×106m
Step 2: Next, we calculate the total distance from the center of the Earth to
the satellite’s position. This is the sum of the Earth’s radius and the altitude
of the satellite:
r= Altitude + Earth’s radius = 6.871 ×106m + 6.371 ×106m = 13.242 ×106m
Step 3: Now, we can calculate the minimum speed needed for the satellite
to remain in orbit by equating the gravitational force between the Earth and
the satellite to the centripetal force required for circular motion:
GMm
r2=mv2
r
Step 4: Simplify the equation by canceling out the mass of the satellite, and
solve for the velocity v:
v=rGM
r
Step 5: Substitute the known values into the equation:
v=r6.67 ×10−11 m3/kg ·s2×5.97 ×1024 kg
13.242 ×106m
v=r39.799 ×1013
13.242 ×106
v=√3003.6
v≈54.81 m/s
Therefore, the minimum speed the satellite must have to remain in orbit is
approximately 54.81 m/s.
Question 20
Question
A satellite is in a circular orbit around the Earth at an altitude where the
acceleration due to gravity is only 2
Solution
Let’s denote the acceleration due to gravity at the Earth’s surface as gand the
acceleration due to gravity at the altitude of the satellite as 0.02g. We know
that the acceleration due to gravity is given by the formula:
g=G·MEarth
R2
Earth
,
where Gis the gravitational constant, MEarth is the mass of the Earth, and
REarth is the radius of the Earth.
Step 1: Express 0.02gin terms of gto find the altitude of the satellite.
Since the acceleration due to gravity at the altitude of the satellite is 0.02g,
we have:
0.02g=G·MEarth
(REarth +h)2,
where his the altitude of the satellite above the surface of the Earth.
Step 2: Substitute the given values into the equation and solve for h.
Substitute g=G·MEarth
R2
Earth
into the equation to get:
0.02 G·MEarth
R2
Earth =G·MEarth
(REarth +h)2.
Solving for h, we obtain:
h=q49R2
Earth −REarth = 7REarth −REarth = 6REarth = 6 ·6.37 ×106m.
Therefore, the altitude of the satellite is 38.22 ×106m.
Question 21
Question
A satellite of mass mis in a circular orbit around a planet of mass M. The
radius of the orbit is R. The satellite travels at a constant speed v. Determine
the gravitational force exerted on the satellite by the planet in terms of m,M,
R, and v.
Solution
Step 1: The centripetal force required to keep an object moving in a circular
path is provided by the gravitational force. Therefore, we have:
Fgravity =Fcentripetal
Step 2: The gravitational force between the satellite and the planet is given
by Newton’s law of universal gravitation:
Fgravity =GmM
R2
Step 3: The centripetal force required for circular motion can be expressed
as:
Fcentripetal =mv2
R
Step 4: Setting Fgravity =Fcentripetal, we have:
GmM
R2=mv2
R
Step 5: Solving for Fgravity, we get:
Fgravity =mv2
R=GmM
R2
Therefore, the gravitational force exerted on the satellite by the planet in
terms of m,M,R, and vis Fgravity =mv2
R.
Question 22
Question
A satellite is in a circular orbit around a planet with a radius of 5,000 km. If
the satellite completes one orbit in 5 hours, determine the mass of the planet.
Solution
Step 1: Find the speed of the satellite. The speed, v, of an object in circular
motion can be calculated using the formula:
v=2πr
T
where ris the radius of the orbit and Tis the time period of one complete orbit.
Substitute r= 5,000 km and T= 5 hours into the formula:
v=2π×5,000
5
v=10,000π
5
v= 2,000πkm/h
Step 2: Use the gravitational force formula. The centripetal force required
to keep the satellite in orbit is provided by the gravitational force between the
satellite and the planet. This can be calculated using the formula:
F=Gm1m2
r2
where Fis the gravitational force, Gis the gravitational constant, m1is the
mass of the planet, m2is the mass of the satellite, and ris the distance between
the two objects. The gravitational force is also equal to the centripetal force:
F=m2v2
r
Equating the two expressions for Fgives:
Gm1m2
r2=m2v2
r
Gm1=v2r
We are interested in finding the mass of the planet, so we will solve for m1:
m1=v2r
G
Substitute v= 2,000πkm/h, r= 5,000 km, and G= 6.67 ×10−11 m3/kg/s2
into the formula:
m1=(2,000π)2×5,000
6.67 ×10−11
m1=4,000,000π2×5,000
6.67 ×10−11
m1=20,000,000π2
6.67 ×10−11 ≈2.39 ×1021 kg
Therefore, the mass of the planet is approximately 2.39 ×1021 kg.
Question 23
Question
A satellite orbits a planet in a circular path with a radius of 2.5×107meters.
If the satellite completes one orbit in 30 hours, what is the acceleration of the
satellite towards the planet? Assume the orbit is close to the surface of the
planet.
(Given: Gravitational constant G= 6.67 ×10−11 m3/kg ·s2, mass of the
planet M= 5.97 ×1024 kg, radius of the planet r= 6.37 ×106meters)
Solution
Step 1: First, we need to find the speed of the satellite. We know that the time
period of one orbit is 30 hours, so the angular velocity, ω, can be calculated as:
ω=2π
T=2π
30 ×60 ×60 ≈5.79 ×10−5s−1
Step 2: The velocity of the satellite in circular motion can be found using
the formula v=rω. Substituting the values, we get:
v= 2.5×107×5.79 ×10−5≈1447.5 m/s
Step 3: Next, we can calculate the centripetal acceleration of the satellite
using the formula a=v2
r. Substituting the values, we get:
a=(1447.5)2
2.5×107≈1.029 m/s2
Therefore, the acceleration of the satellite towards the planet is approxi-
mately 1.029 m/s2.
Question 24
Question
A satellite is in a circular orbit around Earth with a radius of 8000 km. De-
termine the speed of the satellite in its orbit. Given that the mass of Earth is
5.97 ×1024 kg and the gravitational constant is 6.67 ×10−11 N m2/kg2.
Solution
Step 1: The centripetal force required for the satellite to stay in its circular
orbit is provided by the gravitational force between the satellite and Earth.
Therefore, we have:
m·v2
r=G·M·m
r2
where: m= mass of the satellite v= speed of the satellite r= radius of the
satellite’s orbit M= mass of Earth G= gravitational constant
Step 2: We can cancel out the mass of the satellite, m, from both sides of
the equation:
v2
r=G·M
r2
Step 3: Rearranging the equation to solve for the speed, v, we get:
v=rG·M
r
Step 4: Substituting the known values into the equation, we find:
v=s(6.67 ×10−11 N m2/kg2)·(5.97 ×1024 kg)
8000 ×103m
Step 5: Calculating the speed, we get:
v=r4.00239 ×1014
8×106=p50.029875 ×107= 2236.067998 m/s
Therefore, the speed of the satellite in its circular orbit is approximately
2236.07 m/s.
Question 25
Question
A small object of mass mis attached to a string of length Land is swung in a
horizontal circle with a constant speed v. If the string makes an angle θwith
the vertical direction, find the tension in the string in terms of m,L,v, and θ.
Assume no air resistance.
Solution
Step 1: Draw a free-body diagram of the object in circular motion. The forces
acting on the object are the tension Tin the string, the gravitational force
mg pointing downward, and the centripetal force mv2/L pointing towards the
center of the circle.
Step 2: Resolve the forces into their component directions. The gravita-
tional force mg has a vertical component mg cos(θ) and a horizontal component
mg sin(θ). The tension Thas a horizontal component Tcos(θ) and a vertical
component Tsin(θ).
Step 3: Write the force balance equations in the vertical and horizontal direc-
tions: Vertical direction: Tsin(θ) = mg cos(θ) Horizontal direction: Tcos(θ) =
mv2
L
Step 4: Solve the equation Tsin(θ) = mg cos(θ) for T:T=mg cos(θ)
sin(θ)
Step 5: Substitute the expression for Tinto the horizontal force balance
equation: mg cos(θ)
sin(θ)cos(θ) = mv2
L
Step 6: Simplify the equation to solve for T:T=mv2
Lsin(θ)
Therefore, the tension in the string is mv2
Lsin(θ).
Question 26
Question
A satellite is in a circular orbit around Earth at an altitude of 500 km above
the surface. If the satellite’s speed is 7.5 km/s, what is its period of revolution?
Given that the radius of Earth is 6370 km and its mass is 5.97 ×1024 kg.
Solution
Step 1: Calculate the total distance the satellite travels in one revolution.
The total distance can be calculated as the circumference of the circular
orbit. The radius of the orbit is the sum of the radius of Earth and the altitude
of the satellite:
R= 6370 km + 500 km = 6870 km
The circumference is given by 2πR.
Step 2: Calculate the period of revolution using the distance traveled and
the satellite’s speed.
The speed of the satellite is equal to the distance traveled divided by the
time taken:
v=2πR
T
Where: - v= 7.5 km/s (speed of the satellite), - R= 6870 km (radius of the
orbit).
Rearranging the formula to solve for the period of revolution T:
T=2πR
v
Substitute the given values:
T=2π×6870
7.5s
T≈2×3.14159 ×6870
7.5≈3279.37 s
Hence, the period of revolution for the satellite is approximately 3279.37 sec-
onds.
Question 27
Question
A satellite is in a circular orbit around Earth at an altitude of 1000 km above
the surface. If the radius of Earth is 6400 km, calculate the satellite’s orbital
speed. Assume the satellite is very far away from the effects of the Earth’s
atmosphere.
Solution
Step 1: We can start by finding the total distance from the center of the Earth to
the satellite’s orbit. This is the sum of the radius of the Earth and the altitude
of the satellite above the surface.
r= 6400 km + 1000 km = 7400 km
Step 2: We can now apply the gravitational force equation to find the satel-
lite’s orbital speed. The gravitational force is equal to the centripetal force
required to keep the satellite in its orbit.
GMm
74002=mv2
7400
Where: - G= 6.67 ×10−11 N m2/kg2(Gravitational constant) - M= 5.97 ×
1024 kg (Mass of Earth) - m= mass of the satellite (which cancels out) - v=
orbital speed
Step 3: We can simplify the equation and solve for v.
v=rGM
r
v=s(6.67 ×10−11 N m2/kg2)(5.97 ×1024 kg)
7400 km ×103m/km
v=r39.8×1013
7400 ×103
v≈p5.37 ×106m/s
v≈2317 m/s
Therefore, the satellite’s orbital speed is approximately 2317 m/s.
Question 28
Question
A satellite is in an elliptical orbit around the Earth. At the point where it is
closest to the Earth (perigee), the speed of the satellite is 4.5×103m/s. At the
point where it is farthest from the Earth (apogee), the speed is 2.5×103m/s.
Calculate the total mechanical energy of the satellite.
Given: Mass of the Earth, M= 5.97 ×1024 kg
Radius of the Earth, R= 6.37 ×106m
Solution
Step 1: Find the semi-major axis of the orbit using the information provided.
At perigee (closest to the Earth): v1= 4.5×103m/s
At apogee (farthest from the Earth): v2= 2.5×103m/s
Using conservation of angular momentum, we have:
r1·v1=r2·v2
where r1and r2are the distances at perigee and apogee respectively.
Step 2: Find the velocities v1and v2at perigee and apogee. The kinetic
energy of the satellite is given by:
KE =1
2mv2
At perigee:
KE1=1
2mv2
1
At apogee:
KE2=1
2mv2
2
Step 3: Find the potential energy of the satellite at each point. The potential
energy of the satellite at a distance rfrom the center of the Earth is given by:
P E =−GMm
r
where Mis the mass of the Earth, Gis the gravitational constant, and mis the
mass of the satellite.
Step 4: Calculate the total mechanical energy of the satellite. The total
mechanical energy Eof the satellite is the sum of its kinetic and potential
energies:
E=KE +P E
Question 29
Question
A satellite is in a circular orbit around the Earth at a distance of 8000 km from
the center of the Earth. If the satellite completes one orbit in 10 hours, calculate
the mass of the Earth. Given G= 6.67 ×10−11 m3kg−1s−2.
Solution
Step 1: First, calculate the velocity of the satellite in its orbit using the formula
for the speed of an object in circular motion. The centripetal force required to
keep the satellite in orbit is provided by the gravitational force.
v=2πr
T
where: v= velocity of the satellite, r= radius of the orbit, T= time period of
one orbit.
Given r= 8000 km and T= 10 hours, convert rto meters and Tto seconds
before substituting into the formula.
Step 2: Calculate the centripetal acceleration of the satellite.
ac=v2
r
Step 3: Equate the centripetal acceleration to the gravitational acceleration,
and solve for the mass of the Earth using Newton’s law of gravitation.
Fc=Fg
mv2
r=GmM
r2
where: Fc= centripetal force, Fg= gravitational force, m= mass of the satellite,
M= mass of the Earth.
Step 4: Substitute the values of vand rinto the equation and solve for M,
the mass of the Earth.
Question 30
Question
A satellite in a circular orbit around Earth has a period of revolution of 12 hours.
If the radius of the orbit is 3 ×107m, find the mass of the Earth. Assume the
satellite is in a vacuum and G= 6.67 ×10−11 m3kg−1s−2.
Solution
Step 1: First, let’s calculate the speed of the satellite using the formula for the
centripetal force in circular motion:
Fcentripetal =m·v2
r
m·v2
r=G·M·m
r2
v2=G·M
r
v=rG·M
r
where: - Fcentripetal is the centripetal force, - mis the mass of the satellite, -
vis the speed of the satellite, - ris the radius of the satellite’s orbit, - Gis the
gravitational constant, - Mis the mass of the Earth.
Step 2: Now, let’s find the speed of the satellite by plugging in the known
values:
v=s(6.67 ×10−11 m3kg−1s−2)·M
3×107
v=r2.223 ×103·M
107
Step 3: Since the period of revolution is given by T=2πr
v, we can express v
in terms of T:
v=2πr
T
Step 4: Substitute the expression for vback into the previous equation:
r2.223 ×103·M
107=2π·3×107
12 ×60 ×60
2.223 ×103·M
107=2π·3×107
12 ×60 ×602
M=2π·3×107
12×60×60 2
×107
2.223 ×103
M= 6.04 ×1024 kg
Therefore, the mass of the Earth is 6.04 ×1024 kg.
