PHYS 231 - UNIVERSITY PHYSICS I - Circu-
lar Motion and Gravitation Question Bank - Set
3
Question 1
Question
A small satellite of mass mis in a circular orbit around a planet of mass M.
The radius of the orbit is r. The satellite is then given a small boost in speed
so that its new speed is 3
2times its original speed. Determine the new radius of
the satellite’s orbit.
Solution
Let vbe the original speed of the satellite, vnew be the new speed of the satellite,
and rnew be the new radius of the satellite’s orbit. The centripetal force required
for the satellite to stay in orbit is provided by the gravitational force between
the satellite and the planet:
mv2
r=GMm
r2
where Gis the gravitational constant.
Step 1: Determine the original speed of the satellite. Given that the cen-
tripetal force is provided by the gravitational force, we have:
mv2
r=GMm
r2
Solving for vgives:
v=rGM
r
Step 2: Determine the new speed of the satellite. Since the new speed is 3
2
times the original speed, we have:
vnew =3
2v=3
2rGM
r
Step 3: Determine the new radius of the satellite’s orbit. The centripetal
force required for the satellite to stay in orbit with the new speed is provided
by the gravitational force between the satellite and the planet:
m(vnew)2
rnew
=GMm
r2
new
Solving for rnew gives:
rnew =(vnew)2
GM =3
22
r=9
4r
Therefore, the new radius of the satellite’s orbit is 9
4times the original radius.
Question 2
Question
A satellite is in circular orbit around a planet with a period of 12 hours. If the
radius of the orbit is 1.1×107m, determine the mass of the planet. Assume the
satellite’s mass is negligible compared to the planet’s mass.
Solution
Step 1: First, we can find the speed of the satellite using the formula for the
orbital speed of a satellite in circular motion:
v=2πr
T
where: - vis the orbital speed, - ris the radius of the orbit, and - Tis the
period of the orbit.
Substitute the given values:
v=2π×1.1×107
12 ×3600
v=2.2π×107
43200
v= 1618.37 m/s
Step 2: Next, we can use the gravitational force equation to find the mass
of the planet:
F=GMm
r2
where: - Fis the gravitational force, - Gis the universal gravitational constant,
-Mis the mass of the planet, - mis the mass of the satellite, and - ris the
radius of the orbit.
The gravitational force is providing the centripetal force for the satellite’s
circular motion: GMm
r2=mv2
r
Substitute the known values:
GM
r=v2
G×M
1.1×107= 1618.372
G×M= 1618.372×1.1×107
G×M= 29936402606
Step 3: Finally, we solve for M:
M=29936402606
G
M=29936402606
6.674 ×10−11
M≈4.49 ×1017 kg
Therefore, the mass of the planet is approximately 4.49 ×1017 kg.
Question 3
Question
A satellite is in a circular orbit around Earth with a radius of 10,000 km. If
the satellite completes one orbit in 24 hours, what is the magnitude of the
gravitational force acting on the satellite? The mass of Earth is 5.97 ×1024 kg
and the gravitational constant is 6.67 ×10−11 Nm2/kg2.
Solution
Step 1: Calculate the speed of the satellite in its orbit using the formula for the
orbital speed in a circular orbit:
v=2πr
T
where v= orbital speed, r= 10,000 km = 10,000,000 m (radius of orbit),
T= 24 hours = 86,400 s (time taken for one orbit).
Substitute the given values into the formula:
v=2π×10,000,000
86,400
Step 2: Calculate the speed of the satellite in m/s.
v= 7272 m/s
Step 3: Calculate the acceleration of the satellite in its orbit using the for-
mula for centripetal acceleration:
a=v2
r
where v= 7272 m/s (calculated in step 2), r= 10,000,000 m (radius of orbit).
Substitute the given values into the formula:
a=(7272)2
10,000,000
Step 4: Calculate the acceleration of the satellite in m/s2.
a= 52.5 m/s2
Step 5: Calculate the gravitational force acting on the satellite using New-
ton’s law of universal gravitation:
F=m·v2
r
where F= gravitational force, m= 5.97 ×1024 kg (mass of Earth), v= 7272
m/s (orbital speed), r= 10,000,000 m (radius of orbit).
Substitute the given values into the formula:
F=5.97 ×1024 ×(7272)2
10,000,000
Step 6: Calculate the magnitude of the gravitational force acting on the
satellite.
F≈4.26 ×1020 N
Therefore, the magnitude of the gravitational force acting on the satellite is
approximately 4.26 ×1020 N.
Question 4
Question
A satellite of mass mis in a circular orbit around a planet of mass M. The
radius of the orbit is Rand the gravitational force between the satellite and
the planet provides the centripetal force needed to keep the satellite in circular
motion. Determine the speed of the satellite in terms of G,M,m, and R.
Solution
Let vbe the speed of the satellite in its circular orbit. The gravitational force
between the satellite and the planet is given by the formula:
Fgrav =GMm
R2
And the centripetal force required for circular motion is given by:
Fcentripetal =mv2
R
Since the gravitational force provides the centripetal force, we have:
Fgrav =Fcentripetal
Therefore,
GMm
R2=mv2
R
Solving for v, we get:
v=rGM
R
Question 5
Question
A satellite is in a circular orbit around a planet with a radius of 5000 km. If
the period of the satellite’s orbit is 6 hours, calculate the mass of the planet.
Solution
Step 1: Calculate the orbital speed of the satellite using the formula for the
orbital period of a satellite in circular orbit:
v=2πr
T
where v= orbital speed of the satellite, r= radius of the orbit, and T= period
of the orbit.
Substitute r= 5000 km and T= 6 hours into the formula:
v=2π×5000
6
v=10000π
6≈5235.98 km/h
Step 2: Calculate the centripetal acceleration of the satellite using the for-
mula:
ac=v2
r
where ac= centripetal acceleration, v= orbital speed of the satellite, and r=
radius of the orbit.
Substitute v≈5235.98 km/h and r= 5000 km into the formula:
ac=(5235.98)2
5000 ≈5467.06 km/h2
Step 3: Find the gravitational force acting on the satellite using Newton’s
law of universal gravitation:
F=msv2
r
where F= gravitational force, ms= mass of the satellite, v= orbital speed of
the satellite, and r= distance from the center of the planet.
Step 4: Calculate the mass of the planet using the formula for gravitational
force:
F=G·Mp·ms
r2
where G= gravitational constant, Mp= mass of the planet, ms= mass of the
satellite, and r= radius of the orbit.
Now, equate the two expressions for gravitational force and solve for Mp:
msv2
r=G·Mp·ms
r2
Mp=v2
G·r
Substitute v≈5235.98 km/h, G= 6.67 ×10−11 Nm2/kg2, and r= 5000 km
into the formula:
Mp=(5235.98)2
6.67 ×10−11 ·5000 ≈1.22 ×1024 kg
Therefore, the mass of the planet is approximately 1.22 ×1024 kg.
Question 6
Question
A satellite of mass mis orbiting a planet of mass Min a circular orbit of radius
r. If the gravitational force is the only force acting on the satellite, find an
expression for the orbital speed of the satellite in terms of G,M,m, and r.
Solution
Step 1: The gravitational force provides the centripetal force required to keep
the satellite in circular motion. By Newton’s law of gravitation, the gravitational
force is given by:
Fg=GmM
r2
where Gis the gravitational constant, Mis the mass of the planet, mis the
mass of the satellite, and ris the radius of the orbit.
Step 2: The centripetal force required to keep an object in circular motion
is given by:
Fc=mv2
r
where vis the orbital speed of the satellite.
Step 3: Setting the gravitational force equal to the centripetal force, we have:
GmM
r2=mv2
r
Step 4: Solving for v, we get:
v=rGM
r
Therefore, the orbital speed of the satellite is given by v=qGM
r, where G
is the gravitational constant, Mis the mass of the planet, and ris the radius
of the orbit.
Question 7
Question
A satellite is in a circular orbit around Earth at an altitude where the acceler-
ation due to gravity is 0.8 m/s2. If the speed of the satellite is 7000 m/s, what
is the radius of the orbit?
Solution
Step 1: The acceleration due to gravity at a given altitude above the surface of
Earth can be calculated using the formula
g=GM
(R+h)2
where g= 0.8m/s2is the acceleration due to gravity, G= 6.67×10−11 m3/kg s2
is the universal gravitational constant, M= 5.97 ×1024 kg is the mass of Earth,
R= 6371 km is the radius of Earth, and his the altitude of the satellite above
the surface of Earth.
Step 2: By rearranging the formula for g, we can solve for h:
0.8 = 6.67 ×10−11 ×5.97 ×1024
(6371 ×103+h)2
Step 3: Simplifying the equation, we get
h=r6.67 ×10−11 ×5.97 ×1024
0.8−6371 ×103
Step 4: Calculating the altitude h, we find h≈2.37 ×105m.
Step 5: Since the satellite is in a circular orbit, the gravitational force pro-
vides the centripetal force needed for the circular motion:
GMm
(R+h)2=mv2
R+h
where mis the mass of the satellite, and Ris the radius of the orbit.
Step 6: Canceling out mfrom both sides of the equation and substituting
the known values, we get
6.67 ×10−11 ×5.97 ×1024
(6371 ×103+ 2.37 ×105)2=70002
R+ 2.37 ×105
Step 7: Solving for R, we find R≈3.30 ×107m. Therefore, the radius of
the orbit is approximately 3.30 ×107meters.
Question 8
Question
A small object of mass mis placed 2 meters away from a larger object of mass
M. If the small object is released from rest, what is the speed of the small
object just before it collides with the larger object due to gravity? Assume no
other forces are acting on the objects.
Solution
Step 1: Calculate the gravitational force between the two objects. The gravita-
tional force between the two objects is given by Newton’s law of gravitation:
F=GmM
r2
where Fis the force of gravity, Gis the gravitational constant, mis the mass of
the smaller object, Mis the mass of the larger object, and ris the separation
between the two objects.
Step 2: Calculate the work done by gravity as the small object moves from
2 meters to 0 meters away. The work done by gravity is equal to the change in
potential energy, which is given by:
W= ∆U=−GmM
ri
+GmM
rf
where ri= 2 m and rf= 0 m.
Step 3: The work done by gravity will be equal to the kinetic energy of
the object just before collision. The work done by gravity will be converted to
kinetic energy just before the smaller object collides with the larger object:
W=1
2mv2
where vis the speed of the small object just before collision.
Step 4: Equate the work done by gravity to the kinetic energy and solve for
v.
−GmM
2+GmM =1
2mv2
−GMm
2+GMm
2=1
2mv2
0 = 1
2mv2
v2= 0
So, the speed of the small object just before it collides with the larger object
due to gravity is 0 m/s.
Question 9
Question
A satellite of mass mis in a circular orbit around a planet of mass M. The
radius of the orbit is rand the period of the satellite is T. Find an expression
for the gravitational force acting on the satellite in terms of m,M,r, and T.
Solution
Step 1: Determine the centripetal force acting on the satellite. The centripetal
force required to keep the satellite in a circular orbit is provided by the gravi-
tational force between the satellite and the planet. Therefore, we have:
Fcentripetal =mv2
r
where vis the orbital speed of the satellite.
Step 2: Relate the orbital speed to the radius and period. The orbital speed
of the satellite is given by:
v=2πr
T
Step 3: Substitute the expression for vinto the centripetal force equation.
Substitute v=2πr
Tinto Fcentripetal =mv2
r:
Fcentripetal =m2πr
T2
r
Fcentripetal =4π2mr
T2
Step 4: Equate the centripetal force to the gravitational force. Given that
the centripetal force is provided by the gravitational force, we have:
Fgravitational =Fcentripetal
Thus:
Fgravitational =4π2mr
T2
Therefore, the gravitational force acting on the satellite in terms of m,M,
r, and Tis 4π2mr
T2.
Question 10
Question
A small object of mass mis placed at a distance rfrom the center of a large
uniform spherical object of mass Mand radius R, where r > R. Assuming no
external forces are acting on the system, determine an expression for the mini-
mum initial speed the small object must have in order to escape the gravitational
pull of the large object.
Solution
Let’s denote the gravitational constant as G. The gravitational force between
two objects of masses mand Mseparated by a distance dis given by
F=GmM
d2.
Step 1: The gravitational force on the small object when placed at a distance
rfrom the center of the large object is
F=GmM
r2.
Step 2: The centripetal force required for circular motion of the small object
at a distance rfrom the center of the large object is
F=mv2
r.
Step 3: For the object to escape the gravitational pull, the centripetal force
must be equal to the gravitational force. Therefore, we have
GmM
r2=mv2
r.
Step 4: Simplifying the equation above, we get
v=rGM
r.
Step 5: Thus, the minimum initial speed the small object must have in order
to escape the gravitational pull of the large object is qGM
r.
Question 11
Question
A satellite is in a circular orbit around the Earth at an altitude of 500 km above
the surface. The mass of the Earth is 5.97 ×1024 kg and its radius is 6370 km.
Find the orbital speed of the satellite.
Solution
Step 1: Find the total distance from the center of the Earth to the satellite’s
orbit. The distance from the center of the Earth to the satellite’s orbit is the
sum of the Earth’s radius and the satellite’s altitude above the surface.
d= 6370 km + 500 km = 6870 km = 6.87 ×106m
Step 2: Find the gravitational force between the Earth and the satellite. The
gravitational force between the Earth and the satellite is given by Newton’s law
of gravitation:
F=G·MEarth ·msatellite
d2
where Gis the gravitational constant (6.67 ×10−11 N·m2/kg2), MEarth is the
mass of the Earth (5.97×1024 kg), msatellite is the mass of the satellite (assumed
to be negligible compared to the Earth), and dis the distance between the
centers of the Earth and the satellite.
Step 3: Find the centripetal force required to keep the satellite in circular
motion. The centripetal force required to keep the satellite in circular motion
is equal to the gravitational force between the Earth and the satellite:
Fcentripetal =msatellite ·v2
d
where vis the orbital speed of the satellite.
Step 4: Equate the centripetal force to the gravitational force.
msatellite ·v2
d=G·MEarth ·msatellite
d2
Simplify to find the orbital speed v.
Question 12
Question
A satellite is in a circular orbit around a planet with a period of 24 hours. If
the radius of the orbit is 50,000 km, determine the mass of the planet.
Solution
Step 1: First, we find the speed of the satellite in its orbit. The centripetal
acceleration required to keep the satellite in its circular orbit is provided by the
gravitational force:
ac=v2
r=GM
r2
where vis the speed of the satellite, ris the radius of the orbit, Gis the
gravitational constant, and Mis the mass of the planet.
Step 2: Since the time period Tis related to the speed vand distance s
traveled by the satellite by v=s
Tand s= 2πr, we have
v=2πr
T=2π×50,000 km
24 hours ×1 hour
3600 s
Step 3: Substitute the expression for vinto the centripetal acceleration
equation to solve for M:
(2π×50,000 ×1000 m)2
(24 ×3600)2×50,000 ×103=GM
(50,000 ×103)2
Step 4: Rearrange the equation to solve for M:
M=(2π×50,000 ×1000)2×(50,000 ×103)2
(24 ×3600)2×G
Step 5: Calculate the mass of the planet using the above formula.
Question 13
Question
A satellite is orbiting a planet at a speed of 3.5×104m/s. The satellite is in
a circular orbit at an altitude of 500 km above the planet’s surface. Determine
the mass of the planet in kilograms.
Solution
Step 1: First, let’s find the orbital radius of the satellite orbiting the planet.
The altitude above the planet’s surface is 500 km, which should be added to the
radius of the planet to get the total orbital radius. The total orbital radius, r,
can be expressed as:
r= radius of planet + altitude above surface
Step 2: The satellite is in a circular orbit around the planet, so the gravita-
tional force provides the centripetal force necessary for circular motion. The
centripetal force is given by:
Fc=mv2
r
where mis the mass of the satellite, vis the speed of the satellite, and ris the
distance from the satellite to the center of the planet. Step 3: The gravitational
force is given by Newton’s law of universal gravitation:
Fg=Gmpm
r2
where mpis the mass of the planet, mis the mass of the satellite, ris the radius
of the orbit, and Gis the gravitational constant. Step 4: Setting the centripetal
force equal to the gravitational force, and then solving for mp, we get:
mv2
r=Gmpm
r2
mp=v2r
G
Step 5: Plug in the values for v,r, and Gto find the mass of the planet.
Remember to convert the altitude to meters (1 km = 1000 m). Feel free to ask
any questions if you need further clarification or assistance.
Question 14
Question
A particle is in uniform circular motion in a horizontal plane with a radius of
2 meters. If the particle completes one full revolution in 3 seconds, what is the
magnitude of the acceleration of the particle?
Solution
Step 1: First, we find the angular velocity of the particle. The angular velocity,
ω, is given by:
ω=2π
T
where Tis the time taken to complete one full revolution. Given that T= 3
seconds, we have:
ω=2π
3
Step 2: Next, we find the magnitude of the velocity of the particle. The
magnitude of the velocity, v, in uniform circular motion is given by:
v=rω
where ris the radius of the circle. Substituting r= 2 meters and ω=2π
3, we
get:
v= 2 ×2π
3=4π
3m/s
Step 3: Finally, we find the magnitude of the acceleration of the particle.
The acceleration of an object moving in uniform circular motion is given by:
a=rω2
Substituting r= 2 meters and ω=2π
3into the equation, we get:
a= 2 ×2π
32
=4π2
9m/s2
Therefore, the magnitude of the acceleration of the particle is 4π2
9m/s2.
Question 15
Question
A small object of mass mis attached to a string and is whirled in a vertical
circle of radius rat a constant speed v. Calculate the tension in the string at
the highest point of the circle.
Solution
Step 1: At the highest point of the circle, the tension in the string provides the
centripetal force needed to keep the object moving in a circle.
Step 2: The forces acting on the object at the highest point are the tension
in the string (T) pointing downwards, the weight of the object (mg) pointing
downwards, and the centrifugal force (mv2/r) pointing upwards.
Step 3: To find the tension in the string, we set up the equation of forces in
the vertical direction:
T−mg =mv2
r
Step 4: Here, the centrifugal force is equal in magnitude but opposite in
direction to the component of the weight of the object perpendicular to the
circular path.
Step 5: Solving for T, we have:
T=mg +mv2
r
Step 6: Substituting the values of weight and centripetal acceleration, we
get:
T=m(g+v2
r)
Hence, the tension in the string at the highest point of the circle is m(g+v2
r).
Question 16
Question
A satellite is in a circular orbit around a planet with a radius of 5000 km. If
the satellite has a speed of 10 km/s, calculate the mass of the planet.
Solution
Step 1: Start by finding the acceleration of the satellite using the formula for
centripetal acceleration, a=v2
r, where vis the speed of the satellite and ris
the radius of the orbit.
Given: v= 10 km/s = 10000 m/s, r= 5000 km = 5000000 m.
Substitute these values into the formula:
a=(10000)2
5000000 = 20 m/s2
Step 2: Now, use Newton’s law of gravitation, F=GMm
r2, where Fis the
gravitational force between the satellite and the planet, Mis the mass of the
planet, mis the mass of the satellite, ris the radius of the orbit, and Gis the
gravitational constant.
The gravitational force is also equal to ma, where mis the mass of the
satellite and ais the acceleration of the satellite. Therefore, ma =GMm
r2.
Step 3: Cancel out the mass of the satellite, m, on both sides of the equation:
a=GM
r2
Step 4: Substitute the value of acceleration calculated in Step 1 into the
equation:
20 = GM
(5000000)2
Step 5: Rearrange the equation to solve for M:
M=20 ×(5000000)2
G
Step 6: Substitute the value of the gravitational constant, G= 6.67 ×
10−11 N m2/kg2, into the equation:
M=20 ×(5000000)2
6.67 ×10−11
Step 7: Calculate the mass of the planet to find:
M≈7.5×1022 kg
Therefore, the mass of the planet is approximately 7.5×1022 kg.
Question 17
Question
A satellite of mass mis in a circular orbit around a planet of mass Mand radius
R. If the satellite completes one orbit in time T, determine the speed of the
satellite in terms of the given quantities.
Solution
Step 1: The gravitational force between the satellite and the planet provides
the centripetal force for the satellite’s circular motion. Equating these forces,
we have: GMm
R2=mv2
R
where Gis the gravitational constant.
Step 2: Simplifying the equation from Step 1, we find:
v=rGM
R
Step 3: The time Tfor one orbit is related to the speed vand the circum-
ference of the orbit 2πR by:
T=2πR
v
Step 4: Substituting the expression for vfrom Step 2 into the equation from
Step 3, we get:
T=2πR
qGM
R
Step 5: Simplifying the expression in Step 4, we find:
v=rGM
R
Therefore, the speed of the satellite in terms of the given quantities is qGM
R.
Question 18
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 and travels at a speed of 8000
m/s, determine the gravitational force acting on the satellite.
Given: Radius of Earth, R= 6.371×106m Mass of Earth, M= 5.972×1024
kg Universal gravitational constant, G= 6.674 ×10−11 m3kg−1s−2
Solution
Step 1: Calculate the gravitational force on the satellite using Newton’s law of
gravitation: The force of gravity on the satellite is given by:
F=GMm
r2
where: Gis the universal gravitational constant (6.674×10−11 m3kg−1s−2),
Mis the mass of the Earth (5.972×1024 kg), mis the mass of the satellite (1000
kg), ris the distance from the center of the Earth to the satellite’s orbit (radius
of Earth + altitude of satellite).
Given that the radius of the Earth (R) is 6.371 ×106m, and the satellite’s
altitude above the Earth’s surface is 500 km = 500 ×103m, we have: r=
R+ altitude = 6.371 ×106+ 500 ×103
Step 2: Calculate the total distance from the center of the Earth to the
satellite’s orbit:
r= 6.371 ×106+ 500 ×103= 6.371 ×106+ 5 ×105= 6.821 ×106m
Step 3: Substitute the values into the formula and solve for the gravitational
force:
F=(6.674 ×10−11 ×5.972 ×1024 ×1000)
(6.821 ×106)2
Step 4: Calculate the gravitational force:
F=3.986368 ×1014
4.65946841 ×1013 ≈8.55 ×100N
Therefore, the gravitational force acting on the satellite is approximately
8.55 N.
Question 19
Question
A satellite is in a circular orbit around a planet with a radius of 5.0×106m.
If the satellite completes one orbit in 1.5 hours, determine the orbital speed of
the satellite.
Solution
Step 1: We can determine the orbital speed of the satellite using the formula
for the orbital speed of an object in circular motion:
v=2πr
T
where: v= orbital speed, r= radius of the orbit, and T= time period of the
orbit.
Step 2: Given that r= 5.0×106m and T= 1.5 hours, we can substitute
these values into the formula:
v=2π×5.0×106
1.5×3600
Step 3: Simplifying the expression, we have:
v=10π×106
5400
v=10π×106
5400 ×103
103
v=10π×109
5400 ×103
v=10π×109
5.4×106
v=10π
5.4×103
v≈58.52 km/s
Step 4: Therefore, the orbital speed of the satellite is approximately 58.52
km/s.
Question 20
Question
A satellite is in a circular orbit around a planet with a period of 3 hours. If the
radius of the orbit is increased by a factor of 3, what will be the new period of
the satellite?
Solution
Step 1: First, we need to find the initial velocity of the satellite in its original
circular orbit using the formula for the centripetal force:
Fcentripetal =m·v2
r=G·M·m
r2
where Fcentripetal is the centripetal force, mis the mass of the satellite, vis its
velocity, ris the radius of the orbit, Gis the gravitational constant, Mis the
mass of the planet, and mis the mass of the satellite.
Step 2: We know that the centripetal force is also equal to:
Fcentripetal =m·4π2·r
T2
where Tis the period of the satellite’s orbit.
Step 3: Setting these two expressions for the centripetal force equal to each
other, we can solve for v:
G·M
r2=4π2·r
T2
v=rG·M
r
Step 4: Now, for the orbital period to change, we need the centripetal force
to change, which can be achieved by changing the velocity. When we change
the radius by a factor of 3, the velocity must also change by the same factor to
keep the satellite in orbit.
Step 5: Therefore, the new velocity in the larger orbit will be 1
3times the
original velocity:
vnew =1
3·rG·M
3r
Step 6: Finally, we can find the new period Tnew in the larger orbit by using
the centripetal force formula for the larger orbit:
Fcentripetal =m·(vnew)2
3r=4π2·3r
T2
new
G·M
3r2=12π2·r
T2
new
Tnew =r12π2·r3
G·M
Therefore, the new period of the satellite in the larger orbit will be q12π2·r3
G·M.
Question 21
Question
A satellite is in a circular orbit around the Earth at an altitude of 500 km above
the surface. Determine the speed of the satellite in its orbit. The mass of the
Earth is 5.972 ×1024 kg and its radius is 6,371 km.
Solution
Step 1: Calculate the total radius of the satellite’s orbit. The total radius of
the satellite’s orbit is the sum of the radius of the Earth and the altitude of the
satellite.
r= 6371 km + 500 km = 6871 km
Step 2: Calculate the gravitational force acting on the satellite. The gravi-
tational force can be calculated using Newton’s law of universal gravitation:
F=Gm1m2
r2
where G= 6.674×10−11 m3kg−1s−2is the gravitational constant, m1= 5.972×
1024 kg is the mass of the Earth, m2is the mass of the satellite (we can ignore
this compared to the Earth’s mass), and r= 6871×103m is the distance between
the Earth’s center and the satellite.
F=(6.674 ×10−11 m3kg−1s−2)×(5.972 ×1024 kg)
(6871 ×103m)2
F= 8.87 ×106N
Step 3: Calculate the centripetal force required for the circular motion. The
centripetal force required to keep the satellite in a circular orbit is equal to the
gravitational force acting on it:
Fcentripetal =Fgravitational =mv2
r
where mis the mass of the satellite (which we ignored earlier), vis the velocity
of the satellite, and r= 6871 ×103m is the radius of the orbit. Setting the
gravitational force equal to the centripetal force:
mv2
r=Fgravitational
mv2
r= 8.87 ×106N
Step 4: Calculate the satellite’s speed in its orbit. Since the mass of the
satellite cancels out, we can solve for v:
v=rFgravitational ×r
m
v=r8.87 ×106N×6871 ×103m
m
Therefore, the speed of the satellite in its orbit is q8.87×106N×6871×103m
m.
Question 22
Question
A satellite of mass mis in a circular orbit around a planet of mass M. The
satellite is at a distance rfrom the center of the planet. Show that the period
Tof the satellite’s orbit is given by:
T= 2πsr3
G(M+m)
where Gis the gravitational constant.
Solution
Step 1: The force of gravity between the satellite and the planet provides the
centripetal force required for the satellite to move in a circular orbit. Thus, we
have: GMm
r2=mv2
r
where vis the speed of the satellite.
Step 2: From the equation above, we can solve for the speed of the satellite
in terms of M,m, and r:
v=rGM
r
Step 3: The period Tof the satellite’s orbit is the time it takes for the
satellite to complete one full revolution around the planet. It can be calculated
as:
T=2πr
v
Step 4: Substituting the expression for vfrom Step 2 into the equation for
Tin Step 3, we get:
T=2πr
qGM
r
= 2πrr3
GM
Step 5: We can further simplify the expression by replacing GM with G(M+
m) since the total mass acting on the satellite is M+m:
T= 2πsr3
G(M+m)
Therefore, the period Tof the satellite’s orbit around the planet is given by
T= 2πqr3
G(M+m).
Question 23
Question
A satellite is in circular orbit around a planet with a radius of 2.5×107m. The
satellite has a mass of 500 kg. If the speed of the satellite is 1.75 ×104m/s,
find the period of the satellite’s orbit around the planet.
Solution
Step 1: Find the gravitational force acting on the satellite. The gravitational
force acting on the satellite is given by the formula:
Fg=G·m1·m2
r2
where: - G= 6.67×10−11 m3kg−1s−2(gravitational constant), - m1is the mass
of the planet, - m2is the mass of the satellite, and - ris the radius of the orbit.
Substitute the given values into the formula:
Fg=(6.67 ×10−11 m3kg−1s−2)·m1·m2
r2
Step 2: Find the speed of the satellite. The centripetal force required to
keep the satellite in circular motion is equal to the gravitational force:
Fc=Fg
The centripetal force is given by the formula:
Fc=m·v2
r
where mis the mass of the satellite, vis the speed of the satellite, and ris the
radius of the orbit. Set the centripetal force equal to the gravitational force and
solve for speed:
m·v2
r=(6.67 ×10−11 m3kg−1s−2)·m1·m2
r2
Step 3: Find the period of the satellite’s orbit. The period of the satellite’s
orbit can be calculated using the formula:
T=2πr
v
where Tis the period of the orbit, ris the radius of the orbit, and vis the speed
of the satellite. Substitute the known values into the formula to find the period.
Question 24
Question
A satellite is in a circular orbit around a planet. The speed of the satellite is
2.5×104m/s and the radius of the orbit is 1.2×107m. Calculate the mass of
the planet the satellite is orbiting.
Solution
Step 1: Recall the centripetal force required for an object moving in a circular
path:
Fc=mv2
r
where Fcis the centripetal force, mis the mass of the satellite, vis the speed
of the satellite, and ris the radius of the orbit.
Step 2: The gravitational force between the satellite and the planet provides
the centripetal force:
Fc=GmM
r2
where Gis the universal gravitational constant, Mis the mass of the planet,
and ris the radius of the orbit.
Step 3: Set the centripetal force equations equal to each other:
GmM
r2=mv2
r
Step 4: Simplify the equation by canceling out m:
GM =v2r
G
Step 5: Substitute the given values v= 2.5×104m/s and r= 1.2×107m,
and G= 6.67 ×10−11 N m2/kg2:
M=(2.5×104)2×1.2×107
6.67 ×10−11
Step 6: Calculate the mass of the planet:
M=6.25 ×108×1.2×107
6.67 ×10−11 =7.5×1015
6.67 ×10−11
Step 7: Therefore, the mass of the planet is:
M= 1.125 ×1027 kg
Question 25
Question
A satellite is in a circular orbit around a planet of mass M. If the velocity of
the satellite is doubled, what will happen to the radius of the orbit?
Solution
Let v1be the initial velocity of the satellite, r1the initial radius of the orbit, v2
the final velocity of the satellite, and r2the final radius of the orbit.
Step 1: We can start by relating the initial velocity and radius to the grav-
itational force between the satellite and the planet. The centripetal force is
provided by the gravitational force:
GMm
r2
1
=mv2
1
r1
where Gis the gravitational constant, mis the mass of the satellite, and Mis
the mass of the planet.
Step 2: Similarly, the final velocity and radius are related by:
GMm
r2
2
=mv2
2
r2
Step 3: Given that v2= 2v1, we can substitute this into the equation in Step
2 to find a relation between r1and r2.
GMm
r2
2
=m(2v1)2
r2
⇒GM
r2
= 4v2
1
r2
⇒r2=GM
4v2
1
Step 4: Now, we can substitute r1back into the equation in Step 1 to relate
r2to r1:
GM
r2
1
=mv2
1
r1
⇒r1=GM
v2
1
Step 5: Finally, substitute r1into the expression found in Step 4 for r2:
r2=GM
4v2
1
=GM
4GM
v2
1=r1
4
Step 6: Therefore, if the velocity of the satellite is doubled, the radius of the
orbit will be divided by 4.
Question 26
Question
A small object with mass mis tied to a string and whirled around in a horizontal
circle at a constant speed. The string makes an angle θwith the vertical as
shown in the figure. Calculate the tension in the string in terms of m,θ, and
physical constants.
r
mg
Tθ
Solution
Step 1: Identify the forces acting on the object. The forces acting on the object
are the tension Tin the string and the gravitational force mg acting downward.
Step 2: Break the gravitational force into components. The gravitational
force can be broken down into two components: one along the direction of the
string and the other perpendicular to it.
Step 3: Write the equations for the forces along each direction. In the radial
direction, the tension provides the centripetal force:
Tcos θ=mv2
r
In the vertical direction, the forces are balanced:
Tsin θ=mg
Step 4: Solve for Tin terms of m,θ, and physical constants. Divide the
equation for the radial direction by the equation for the vertical direction:
Tcos θ
Tsin θ=mv2
mgr
Simplify to solve for T:
tan θ=v2
gr
v=pgr tan θ
Substitute this into the equation for the tension in the vertical direction:
Tsin θ=mg
T=mg
sin θ
Therefore, the tension in the string is given by T=mg
sin θ.
Question 27
Question
A planet of mass 2.0×1024 kg and radius 6.0×106m rotates about an axis
perpendicular to the plane of its equator. The planet’s gravitational field pro-
duces a weight of 800 N for a body on the equator. What is the planet’s period
of rotation?
Solution
Step 1: Calculate the acceleration due to gravity on the planet’s surface using
the weight provided. Step 2: Use the formula for centripetal acceleration to find
the planet’s angular velocity. Step 3: Calculate the planet’s period of rotation
using the angular velocity obtained.
Step 1: The weight Wof the body on the equator is given as 800 N. The
weight of an object is given by W=mg, where mis the mass of the object and
gis the acceleration due to gravity.
Given: Mass of the planet, mp= 2.0×1024 kg Radius of the planet, r=
6.0×106m Weight of the body, W= 800 N
The acceleration due to gravity gon the planet’s surface is:
g=W
m
g=800
m
Step 2: The centripetal acceleration acof an object moving in a circle of
radius rwith constant speed vis given by:
ac=v2
r
In circular motion, the centripetal acceleration is provided by the gravita-
tional force:
ac=GM
r2
Equating the two expressions for ac, we have:
GM
r2=v2
r
Solving for vgives:
v=rGM
r
Step 3: The period Tof rotation is given by:
T=2πr
v
Substitute the expression for vinto the equation for T:
T=2πr
qGM
r
Since GM =gr2:
T= 2πsr3
g
Now, plug in the given values to find the planet’s period of rotation.
Question 28
Question
A small object of mass mis attached to a string and moves in a vertical circle
of radius rat a constant speed. The string makes an angle θwith the vertical.
What is the tension in the string at the highest point of the circle?
Solution
1. At the highest point of the circle, the tension in the string provides the
centripetal force required to keep the object moving in a circle. The forces
acting on the object are the tension force T, the gravitational force mg, and the
normal force N. 2. Resolving forces vertically, we have Ncos θ−mg = 0. 3.
Resolving forces horizontally, we have Nsin θ=T. 4. The centripetal force at
the highest point is equal to the tension in the string, which is equal to mv2/r.
5. Combining the equations:
Ncos θ−mg = 0
Nsin θ=mv2
r
6. Solving for Nfrom the first equation gives N=mg/ cos θ. Substituting this
into the second equation gives:
mg
cos θsin θ=mv2
r
7. Simplifying, we get gtan θ=v2/r. Since the speed vis constant, the tension
at the highest point is independent of speed. 8. Therefore, the tension in the
string at the highest point of the circle is T=mg/ cos θ.
Question 29
Question
A satellite is in circular orbit around Earth at an altitude of 500 km. Calculate
the speed of the satellite in its orbit. Assume the radius of Earth is 6400 km.
Solution
Step 1: First, let’s calculate the total distance from the center of Earth to the
center of the satellite’s orbit: Given the altitude of the satellite is 500 km and
the radius of Earth is 6400 km, the total distance is:
6400 km + 500 km = 6900 km
Step 2: Next, we calculate the total acceleration due to gravity at this
distance using the universal law of gravitation:
F=G·M·m
r2
where Fis the force of gravity, Gis the gravitational constant, Mis Earth’s
mass, mis the satellite’s mass, and ris the total distance from the center of
Earth to the center of the satellite’s orbit.
Step 3: Now, we find the gravitational force acting on the satellite:
F=G·M·m
6900 km2
Step 4: In a circular orbit, the force of gravity provides the centripetal force:
m·v2
r=G·M·m
6900 km2
Step 5: Simplify the equation:
v=rG·M
r
Step 6: Substitute the values of G,M, and rto find the speed of the satellite:
v=s6.67 ×10−11 N m2/kg2·5.97 ×1024 kg
6900 km ·103m/km
Step 7: Calculate the speed of the satellite:
v≈p9.81 ×106
v≈3129 m/s
Therefore, the speed of the satellite in its orbit around Earth is approxi-
mately 3129 m/s.
Question 30
Question
A satellite of mass mis in circular orbit around a planet of mass M. The satellite
orbits at a distance rfrom the planet’s center. If the gravitational force between
the satellite and the planet is the only force acting on the satellite, show that
the satellite’s orbital period Tis given by
T= 2πsr3
G(M+m)
where Gis the gravitational constant.
Solution
Step 1: The gravitational force between the satellite and the planet is given by
Newton’s law of universal gravitation:
F=GMm
r2
where Fis the gravitational force, Gis the gravitational constant, and Mand
mare the masses of the planet and satellite respectively.
Step 2: The centripetal force required to keep the satellite in circular motion
is provided by the gravitational force. This centripetal force is given by:
F=mv2
r
where vis the orbital speed of the satellite.
Step 3: Equating the gravitational force and the centripetal force, we have:
GMm
r2=mv2
r
Step 4: Rearranging the equation above to solve for the orbital speed vgives:
v2=GM
r
Step 5: The orbital speed vis related to the orbital period Tand the orbit
circumference 2πr by:
v=2πr
T
Step 6: Substituting the expression for v2from Step 4 into the equation in
Step 5 gives:
2πr
T2
=GM
r
Question 10
Question
A small object of mass mis placed at a distance rfrom the center of a large
uniform spherical object of mass Mand radius R, where r > R. Assuming no
external forces are acting on the system, determine an expression for the mini-
mum initial speed the small object must have in order to escape the gravitational
pull of the large object.
Solution
Let’s denote the gravitational constant as G. The gravitational force between
two objects of masses mand Mseparated by a distance dis given by
F=GmM
d2.
Step 1: The gravitational force on the small object when placed at a distance
rfrom the center of the large object is
F=GmM
r2.
Step 2: The centripetal force required for circular motion of the small object
at a distance rfrom the center of the large object is
F=mv2
r.
Step 3: For the object to escape the gravitational pull, the centripetal force
must be equal to the gravitational force. Therefore, we have
GmM
r2=mv2
r.
Step 4: Simplifying the equation above, we get
v=rGM
r.
Step 5: Thus, the minimum initial speed the small object must have in order
to escape the gravitational pull of the large object is qGM
r.
Question 11
Question
A satellite is in a circular orbit around the Earth at an altitude of 500 km above
the surface. The mass of the Earth is 5.97 ×1024 kg and its radius is 6370 km.
Find the orbital speed of the satellite.
Solution
Step 1: Find the total distance from the center of the Earth to the satellite’s
orbit. The distance from the center of the Earth to the satellite’s orbit is the
sum of the Earth’s radius and the satellite’s altitude above the surface.
d= 6370 km + 500 km = 6870 km = 6.87 ×106m
Step 2: Find the gravitational force between the Earth and the satellite. The
gravitational force between the Earth and the satellite is given by Newton’s law
of gravitation:
F=G·MEarth ·msatellite
d2
where Gis the gravitational constant (6.67 ×10−11 N·m2/kg2), MEarth is the
mass of the Earth (5.97×1024 kg), msatellite is the mass of the satellite (assumed
to be negligible compared to the Earth), and dis the distance between the
centers of the Earth and the satellite.
Step 3: Find the centripetal force required to keep the satellite in circular
motion. The centripetal force required to keep the satellite in circular motion
is equal to the gravitational force between the Earth and the satellite:
Fcentripetal =msatellite ·v2
d
where vis the orbital speed of the satellite.
Step 4: Equate the centripetal force to the gravitational force.
msatellite ·v2
d=G·MEarth ·msatellite
d2
Simplify to find the orbital speed v.
Question 12
Question
A satellite is in a circular orbit around a planet with a period of 24 hours. If
the radius of the orbit is 50,000 km, determine the mass of the planet.
Solution
Step 1: First, we find the speed of the satellite in its orbit. The centripetal
acceleration required to keep the satellite in its circular orbit is provided by the
gravitational force:
ac=v2
r=GM
r2
where vis the speed of the satellite, ris the radius of the orbit, Gis the
gravitational constant, and Mis the mass of the planet.
Step 2: Since the time period Tis related to the speed vand distance s
traveled by the satellite by v=s
Tand s= 2πr, we have
v=2πr
T=2π×50,000 km
24 hours ×1 hour
3600 s
Step 3: Substitute the expression for vinto the centripetal acceleration
equation to solve for M:
(2π×50,000 ×1000 m)2
(24 ×3600)2×50,000 ×103=GM
(50,000 ×103)2
Step 4: Rearrange the equation to solve for M:
M=(2π×50,000 ×1000)2×(50,000 ×103)2
(24 ×3600)2×G
Step 5: Calculate the mass of the planet using the above formula.
Question 13
Question
A satellite is orbiting a planet at a speed of 3.5×104m/s. The satellite is in
a circular orbit at an altitude of 500 km above the planet’s surface. Determine
the mass of the planet in kilograms.
Solution
Step 1: First, let’s find the orbital radius of the satellite orbiting the planet.
The altitude above the planet’s surface is 500 km, which should be added to the
radius of the planet to get the total orbital radius. The total orbital radius, r,
can be expressed as:
r= radius of planet + altitude above surface
Step 2: The satellite is in a circular orbit around the planet, so the gravita-
tional force provides the centripetal force necessary for circular motion. The
centripetal force is given by:
Fc=mv2
r
where mis the mass of the satellite, vis the speed of the satellite, and ris the
distance from the satellite to the center of the planet. Step 3: The gravitational
force is given by Newton’s law of universal gravitation:
Fg=Gmpm
r2
where mpis the mass of the planet, mis the mass of the satellite, ris the radius
of the orbit, and Gis the gravitational constant. Step 4: Setting the centripetal
force equal to the gravitational force, and then solving for mp, we get:
mv2
r=Gmpm
r2
mp=v2r
G
Step 5: Plug in the values for v,r, and Gto find the mass of the planet.
Remember to convert the altitude to meters (1 km = 1000 m). Feel free to ask
any questions if you need further clarification or assistance.
Question 14
Question
A particle is in uniform circular motion in a horizontal plane with a radius of
2 meters. If the particle completes one full revolution in 3 seconds, what is the
magnitude of the acceleration of the particle?
Solution
Step 1: First, we find the angular velocity of the particle. The angular velocity,
ω, is given by:
ω=2π
T
where Tis the time taken to complete one full revolution. Given that T= 3
seconds, we have:
ω=2π
3
Step 2: Next, we find the magnitude of the velocity of the particle. The
magnitude of the velocity, v, in uniform circular motion is given by:
v=rω
where ris the radius of the circle. Substituting r= 2 meters and ω=2π
3, we
get:
v= 2 ×2π
3=4π
3m/s
Step 3: Finally, we find the magnitude of the acceleration of the particle.
The acceleration of an object moving in uniform circular motion is given by:
a=rω2
Substituting r= 2 meters and ω=2π
3into the equation, we get:
a= 2 ×2π
32
=4π2
9m/s2
Therefore, the magnitude of the acceleration of the particle is 4π2
9m/s2.
Question 15
Question
A small object of mass mis attached to a string and is whirled in a vertical
circle of radius rat a constant speed v. Calculate the tension in the string at
the highest point of the circle.
Solution
Step 1: At the highest point of the circle, the tension in the string provides the
centripetal force needed to keep the object moving in a circle.
Step 2: The forces acting on the object at the highest point are the tension
in the string (T) pointing downwards, the weight of the object (mg) pointing
downwards, and the centrifugal force (mv2/r) pointing upwards.
Step 3: To find the tension in the string, we set up the equation of forces in
the vertical direction:
T−mg =mv2
r
Step 4: Here, the centrifugal force is equal in magnitude but opposite in
direction to the component of the weight of the object perpendicular to the
circular path.
Step 5: Solving for T, we have:
T=mg +mv2
r
Step 6: Substituting the values of weight and centripetal acceleration, we
get:
T=m(g+v2
r)
Hence, the tension in the string at the highest point of the circle is m(g+v2
r).
Question 16
Question
A satellite is in a circular orbit around a planet with a radius of 5000 km. If
the satellite has a speed of 10 km/s, calculate the mass of the planet.
Solution
Step 1: Start by finding the acceleration of the satellite using the formula for
centripetal acceleration, a=v2
r, where vis the speed of the satellite and ris
the radius of the orbit.
Given: v= 10 km/s = 10000 m/s, r= 5000 km = 5000000 m.
Substitute these values into the formula:
a=(10000)2
5000000 = 20 m/s2
Step 2: Now, use Newton’s law of gravitation, F=GMm
r2, where Fis the
gravitational force between the satellite and the planet, Mis the mass of the
planet, mis the mass of the satellite, ris the radius of the orbit, and Gis the
gravitational constant.
The gravitational force is also equal to ma, where mis the mass of the
satellite and ais the acceleration of the satellite. Therefore, ma =GMm
r2.
Step 3: Cancel out the mass of the satellite, m, on both sides of the equation:
a=GM
r2
Step 4: Substitute the value of acceleration calculated in Step 1 into the
equation:
20 = GM
(5000000)2
Step 5: Rearrange the equation to solve for M:
M=20 ×(5000000)2
G
Step 6: Substitute the value of the gravitational constant, G= 6.67 ×
10−11 N m2/kg2, into the equation:
M=20 ×(5000000)2
6.67 ×10−11
Step 7: Calculate the mass of the planet to find:
M≈7.5×1022 kg
Therefore, the mass of the planet is approximately 7.5×1022 kg.
Question 17
Question
A satellite of mass mis in a circular orbit around a planet of mass Mand radius
R. If the satellite completes one orbit in time T, determine the speed of the
satellite in terms of the given quantities.
Solution
Step 1: The gravitational force between the satellite and the planet provides
the centripetal force for the satellite’s circular motion. Equating these forces,
we have: GMm
R2=mv2
R
where Gis the gravitational constant.
Step 2: Simplifying the equation from Step 1, we find:
v=rGM
R
Step 3: The time Tfor one orbit is related to the speed vand the circum-
ference of the orbit 2πR by:
T=2πR
v
Step 4: Substituting the expression for vfrom Step 2 into the equation from
Step 3, we get:
T=2πR
qGM
R
Step 5: Simplifying the expression in Step 4, we find:
v=rGM
R
Therefore, the speed of the satellite in terms of the given quantities is qGM
R.
Question 18
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 and travels at a speed of 8000
m/s, determine the gravitational force acting on the satellite.
Given: Radius of Earth, R= 6.371×106m Mass of Earth, M= 5.972×1024
kg Universal gravitational constant, G= 6.674 ×10−11 m3kg−1s−2
Solution
Step 1: Calculate the gravitational force on the satellite using Newton’s law of
gravitation: The force of gravity on the satellite is given by:
F=GMm
r2
where: Gis the universal gravitational constant (6.674×10−11 m3kg−1s−2),
Mis the mass of the Earth (5.972×1024 kg), mis the mass of the satellite (1000
kg), ris the distance from the center of the Earth to the satellite’s orbit (radius
of Earth + altitude of satellite).
Given that the radius of the Earth (R) is 6.371 ×106m, and the satellite’s
altitude above the Earth’s surface is 500 km = 500 ×103m, we have: r=
R+ altitude = 6.371 ×106+ 500 ×103
Step 2: Calculate the total distance from the center of the Earth to the
satellite’s orbit:
r= 6.371 ×106+ 500 ×103= 6.371 ×106+ 5 ×105= 6.821 ×106m
Step 3: Substitute the values into the formula and solve for the gravitational
force:
F=(6.674 ×10−11 ×5.972 ×1024 ×1000)
(6.821 ×106)2
Step 4: Calculate the gravitational force:
F=3.986368 ×1014
4.65946841 ×1013 ≈8.55 ×100N
Therefore, the gravitational force acting on the satellite is approximately
8.55 N.
Question 19
Question
A satellite is in a circular orbit around a planet with a radius of 5.0×106m.
If the satellite completes one orbit in 1.5 hours, determine the orbital speed of
the satellite.
Solution
Step 1: We can determine the orbital speed of the satellite using the formula
for the orbital speed of an object in circular motion:
v=2πr
T
where: v= orbital speed, r= radius of the orbit, and T= time period of the
orbit.
Step 2: Given that r= 5.0×106m and T= 1.5 hours, we can substitute
these values into the formula:
v=2π×5.0×106
1.5×3600
Step 3: Simplifying the expression, we have:
v=10π×106
5400
v=10π×106
5400 ×103
103
v=10π×109
5400 ×103
v=10π×109
5.4×106
v=10π
5.4×103
v≈58.52 km/s
Step 4: Therefore, the orbital speed of the satellite is approximately 58.52
km/s.
Question 20
Question
A satellite is in a circular orbit around a planet with a period of 3 hours. If the
radius of the orbit is increased by a factor of 3, what will be the new period of
the satellite?
Solution
Step 1: First, we need to find the initial velocity of the satellite in its original
circular orbit using the formula for the centripetal force:
Fcentripetal =m·v2
r=G·M·m
r2
where Fcentripetal is the centripetal force, mis the mass of the satellite, vis its
velocity, ris the radius of the orbit, Gis the gravitational constant, Mis the
mass of the planet, and mis the mass of the satellite.
Step 2: We know that the centripetal force is also equal to:
Fcentripetal =m·4π2·r
T2
where Tis the period of the satellite’s orbit.
Step 3: Setting these two expressions for the centripetal force equal to each
other, we can solve for v:
G·M
r2=4π2·r
T2
v=rG·M
r
Step 4: Now, for the orbital period to change, we need the centripetal force
to change, which can be achieved by changing the velocity. When we change
the radius by a factor of 3, the velocity must also change by the same factor to
keep the satellite in orbit.
Step 5: Therefore, the new velocity in the larger orbit will be 1
3times the
original velocity:
vnew =1
3·rG·M
3r
Step 6: Finally, we can find the new period Tnew in the larger orbit by using
the centripetal force formula for the larger orbit:
Fcentripetal =m·(vnew)2
3r=4π2·3r
T2
new
G·M
3r2=12π2·r
T2
new
Tnew =r12π2·r3
G·M
Therefore, the new period of the satellite in the larger orbit will be q12π2·r3
G·M.
Question 21
Question
A satellite is in a circular orbit around the Earth at an altitude of 500 km above
the surface. Determine the speed of the satellite in its orbit. The mass of the
Earth is 5.972 ×1024 kg and its radius is 6,371 km.
Solution
Step 1: Calculate the total radius of the satellite’s orbit. The total radius of
the satellite’s orbit is the sum of the radius of the Earth and the altitude of the
satellite.
r= 6371 km + 500 km = 6871 km
Step 2: Calculate the gravitational force acting on the satellite. The gravi-
tational force can be calculated using Newton’s law of universal gravitation:
F=Gm1m2
r2
where G= 6.674×10−11 m3kg−1s−2is the gravitational constant, m1= 5.972×
1024 kg is the mass of the Earth, m2is the mass of the satellite (we can ignore
this compared to the Earth’s mass), and r= 6871×103m is the distance between
the Earth’s center and the satellite.
F=(6.674 ×10−11 m3kg−1s−2)×(5.972 ×1024 kg)
(6871 ×103m)2
F= 8.87 ×106N
Step 3: Calculate the centripetal force required for the circular motion. The
centripetal force required to keep the satellite in a circular orbit is equal to the
gravitational force acting on it:
Fcentripetal =Fgravitational =mv2
r
where mis the mass of the satellite (which we ignored earlier), vis the velocity
of the satellite, and r= 6871 ×103m is the radius of the orbit. Setting the
gravitational force equal to the centripetal force:
mv2
r=Fgravitational
mv2
r= 8.87 ×106N
Step 4: Calculate the satellite’s speed in its orbit. Since the mass of the
satellite cancels out, we can solve for v:
v=rFgravitational ×r
m
v=r8.87 ×106N×6871 ×103m
m
Therefore, the speed of the satellite in its orbit is q8.87×106N×6871×103m
m.
Question 22
Question
A satellite of mass mis in a circular orbit around a planet of mass M. The
satellite is at a distance rfrom the center of the planet. Show that the period
Tof the satellite’s orbit is given by:
T= 2πsr3
G(M+m)
where Gis the gravitational constant.
Solution
Step 1: The force of gravity between the satellite and the planet provides the
centripetal force required for the satellite to move in a circular orbit. Thus, we
have: GMm
r2=mv2
r
where vis the speed of the satellite.
Step 2: From the equation above, we can solve for the speed of the satellite
in terms of M,m, and r:
v=rGM
r
Step 3: The period Tof the satellite’s orbit is the time it takes for the
satellite to complete one full revolution around the planet. It can be calculated
as:
T=2πr
v
Step 4: Substituting the expression for vfrom Step 2 into the equation for
Tin Step 3, we get:
T=2πr
qGM
r
= 2πrr3
GM
Step 5: We can further simplify the expression by replacing GM with G(M+
m) since the total mass acting on the satellite is M+m:
T= 2πsr3
G(M+m)
Therefore, the period Tof the satellite’s orbit around the planet is given by
T= 2πqr3
G(M+m).
Question 23
Question
A satellite is in circular orbit around a planet with a radius of 2.5×107m. The
satellite has a mass of 500 kg. If the speed of the satellite is 1.75 ×104m/s,
find the period of the satellite’s orbit around the planet.
Solution
Step 1: Find the gravitational force acting on the satellite. The gravitational
force acting on the satellite is given by the formula:
Fg=G·m1·m2
r2
where: - G= 6.67×10−11 m3kg−1s−2(gravitational constant), - m1is the mass
of the planet, - m2is the mass of the satellite, and - ris the radius of the orbit.
Substitute the given values into the formula:
Fg=(6.67 ×10−11 m3kg−1s−2)·m1·m2
r2
Step 2: Find the speed of the satellite. The centripetal force required to
keep the satellite in circular motion is equal to the gravitational force:
Fc=Fg
The centripetal force is given by the formula:
Fc=m·v2
r
where mis the mass of the satellite, vis the speed of the satellite, and ris the
radius of the orbit. Set the centripetal force equal to the gravitational force and
solve for speed:
m·v2
r=(6.67 ×10−11 m3kg−1s−2)·m1·m2
r2
Step 3: Find the period of the satellite’s orbit. The period of the satellite’s
orbit can be calculated using the formula:
T=2πr
v
where Tis the period of the orbit, ris the radius of the orbit, and vis the speed
of the satellite. Substitute the known values into the formula to find the period.
Question 24
Question
A satellite is in a circular orbit around a planet. The speed of the satellite is
2.5×104m/s and the radius of the orbit is 1.2×107m. Calculate the mass of
the planet the satellite is orbiting.
Solution
Step 1: Recall the centripetal force required for an object moving in a circular
path:
Fc=mv2
r
where Fcis the centripetal force, mis the mass of the satellite, vis the speed
of the satellite, and ris the radius of the orbit.
Step 2: The gravitational force between the satellite and the planet provides
the centripetal force:
Fc=GmM
r2
where Gis the universal gravitational constant, Mis the mass of the planet,
and ris the radius of the orbit.
Step 3: Set the centripetal force equations equal to each other:
GmM
r2=mv2
r
Step 4: Simplify the equation by canceling out m:
GM =v2r
G
Step 5: Substitute the given values v= 2.5×104m/s and r= 1.2×107m,
and G= 6.67 ×10−11 N m2/kg2:
M=(2.5×104)2×1.2×107
6.67 ×10−11
Step 6: Calculate the mass of the planet:
M=6.25 ×108×1.2×107
6.67 ×10−11 =7.5×1015
6.67 ×10−11
Step 7: Therefore, the mass of the planet is:
M= 1.125 ×1027 kg
Question 25
Question
A satellite is in a circular orbit around a planet of mass M. If the velocity of
the satellite is doubled, what will happen to the radius of the orbit?
Solution
Let v1be the initial velocity of the satellite, r1the initial radius of the orbit, v2
the final velocity of the satellite, and r2the final radius of the orbit.
Step 1: We can start by relating the initial velocity and radius to the grav-
itational force between the satellite and the planet. The centripetal force is
provided by the gravitational force:
GMm
r2
1
=mv2
1
r1
where Gis the gravitational constant, mis the mass of the satellite, and Mis
the mass of the planet.
Step 2: Similarly, the final velocity and radius are related by:
GMm
r2
2
=mv2
2
r2
Step 3: Given that v2= 2v1, we can substitute this into the equation in Step
2 to find a relation between r1and r2.
GMm
r2
2
=m(2v1)2
r2
⇒GM
r2
= 4v2
1
r2
⇒r2=GM
4v2
1
Step 4: Now, we can substitute r1back into the equation in Step 1 to relate
r2to r1:
GM
r2
1
=mv2
1
r1
⇒r1=GM
v2
1
Step 5: Finally, substitute r1into the expression found in Step 4 for r2:
r2=GM
4v2
1
=GM
4GM
v2
1=r1
4
Step 6: Therefore, if the velocity of the satellite is doubled, the radius of the
orbit will be divided by 4.
Question 26
Question
A small object with mass mis tied to a string and whirled around in a horizontal
circle at a constant speed. The string makes an angle θwith the vertical as
shown in the figure. Calculate the tension in the string in terms of m,θ, and
physical constants.
r
mg
Tθ
Solution
Step 1: Identify the forces acting on the object. The forces acting on the object
are the tension Tin the string and the gravitational force mg acting downward.
Step 2: Break the gravitational force into components. The gravitational
force can be broken down into two components: one along the direction of the
string and the other perpendicular to it.
Step 3: Write the equations for the forces along each direction. In the radial
direction, the tension provides the centripetal force:
Tcos θ=mv2
r
In the vertical direction, the forces are balanced:
Tsin θ=mg
Step 4: Solve for Tin terms of m,θ, and physical constants. Divide the
equation for the radial direction by the equation for the vertical direction:
Tcos θ
Tsin θ=mv2
mgr
Simplify to solve for T:
tan θ=v2
gr
v=pgr tan θ
Substitute this into the equation for the tension in the vertical direction:
Tsin θ=mg
T=mg
sin θ
Therefore, the tension in the string is given by T=mg
sin θ.
Question 27
Question
A planet of mass 2.0×1024 kg and radius 6.0×106m rotates about an axis
perpendicular to the plane of its equator. The planet’s gravitational field pro-
duces a weight of 800 N for a body on the equator. What is the planet’s period
of rotation?
Solution
Step 1: Calculate the acceleration due to gravity on the planet’s surface using
the weight provided. Step 2: Use the formula for centripetal acceleration to find
the planet’s angular velocity. Step 3: Calculate the planet’s period of rotation
using the angular velocity obtained.
Step 1: The weight Wof the body on the equator is given as 800 N. The
weight of an object is given by W=mg, where mis the mass of the object and
gis the acceleration due to gravity.
Given: Mass of the planet, mp= 2.0×1024 kg Radius of the planet, r=
6.0×106m Weight of the body, W= 800 N
The acceleration due to gravity gon the planet’s surface is:
g=W
m
g=800
m
Step 2: The centripetal acceleration acof an object moving in a circle of
radius rwith constant speed vis given by:
ac=v2
r
In circular motion, the centripetal acceleration is provided by the gravita-
tional force:
ac=GM
r2
Equating the two expressions for ac, we have:
GM
r2=v2
r
Solving for vgives:
v=rGM
r
Step 3: The period Tof rotation is given by:
T=2πr
v
Substitute the expression for vinto the equation for T:
T=2πr
qGM
r
Since GM =gr2:
T= 2πsr3
g
Now, plug in the given values to find the planet’s period of rotation.
Question 28
Question
A small object of mass mis attached to a string and moves in a vertical circle
of radius rat a constant speed. The string makes an angle θwith the vertical.
What is the tension in the string at the highest point of the circle?
Solution
1. At the highest point of the circle, the tension in the string provides the
centripetal force required to keep the object moving in a circle. The forces
acting on the object are the tension force T, the gravitational force mg, and the
normal force N. 2. Resolving forces vertically, we have Ncos θ−mg = 0. 3.
Resolving forces horizontally, we have Nsin θ=T. 4. The centripetal force at
the highest point is equal to the tension in the string, which is equal to mv2/r.
5. Combining the equations:
Ncos θ−mg = 0
Nsin θ=mv2
r
6. Solving for Nfrom the first equation gives N=mg/ cos θ. Substituting this
into the second equation gives:
mg
cos θsin θ=mv2
r
7. Simplifying, we get gtan θ=v2/r. Since the speed vis constant, the tension
at the highest point is independent of speed. 8. Therefore, the tension in the
string at the highest point of the circle is T=mg/ cos θ.
Question 29
Question
A satellite is in circular orbit around Earth at an altitude of 500 km. Calculate
the speed of the satellite in its orbit. Assume the radius of Earth is 6400 km.
Solution
Step 1: First, let’s calculate the total distance from the center of Earth to the
center of the satellite’s orbit: Given the altitude of the satellite is 500 km and
the radius of Earth is 6400 km, the total distance is:
6400 km + 500 km = 6900 km
Step 2: Next, we calculate the total acceleration due to gravity at this
distance using the universal law of gravitation:
F=G·M·m
r2
where Fis the force of gravity, Gis the gravitational constant, Mis Earth’s
mass, mis the satellite’s mass, and ris the total distance from the center of
Earth to the center of the satellite’s orbit.
Step 3: Now, we find the gravitational force acting on the satellite:
F=G·M·m
6900 km2
Step 4: In a circular orbit, the force of gravity provides the centripetal force:
m·v2
r=G·M·m
6900 km2
Step 5: Simplify the equation:
v=rG·M
r
Step 6: Substitute the values of G,M, and rto find the speed of the satellite:
v=s6.67 ×10−11 N m2/kg2·5.97 ×1024 kg
6900 km ·103m/km
Step 7: Calculate the speed of the satellite:
v≈p9.81 ×106
v≈3129 m/s
Therefore, the speed of the satellite in its orbit around Earth is approxi-
mately 3129 m/s.
Question 30
Question
A satellite of mass mis in circular orbit around a planet of mass M. The satellite
orbits at a distance rfrom the planet’s center. If the gravitational force between
the satellite and the planet is the only force acting on the satellite, show that
the satellite’s orbital period Tis given by
T= 2πsr3
G(M+m)
where Gis the gravitational constant.
Solution
Step 1: The gravitational force between the satellite and the planet is given by
Newton’s law of universal gravitation:
F=GMm
r2
where Fis the gravitational force, Gis the gravitational constant, and Mand
mare the masses of the planet and satellite respectively.
Step 2: The centripetal force required to keep the satellite in circular motion
is provided by the gravitational force. This centripetal force is given by:
F=mv2
r
where vis the orbital speed of the satellite.
Step 3: Equating the gravitational force and the centripetal force, we have:
GMm
r2=mv2
r
Step 4: Rearranging the equation above to solve for the orbital speed vgives:
v2=GM
r
Step 5: The orbital speed vis related to the orbital period Tand the orbit
circumference 2πr by:
v=2πr
T
Step 6: Substituting the expression for v2from Step 4 into the equation in
Step 5 gives:
2πr
T2
=GM
r
Step 7: Solving for the orbital period T, we get:
T= 2πsr3
G(M+m)
Therefore, the satellite’s orbital period Tis given by T= 2πqr3
G(M+m).
