PHYS 231 - UNIVERSITY PHYSICS I - Circular Motion and Gravitation Question Bank - Set 1

PHYS 231 - UNIVERSITY PHYSICS I - Circu-

lar Motion and Gravitation Question Bank - Set

Question 1

Question

A small object with mass mis attached to a string of length Land is swung in

a horizontal circle with a constant speed v. The string makes an angle θwith

the vertical. If the tension in the string is given by T, determine an expression

for the tension Tin terms of m,L,v, and θ.

Solution

Step 1: Draw a free body diagram of the object in circular motion.

ΣFradial =maradial

Step 2: Resolve the forces into radial components. The forces acting on the

object are the tension Tand the component of the object’s weight perpendicular

to the motion.

ΣFradial =Tsin θ−mg cos θ=maradial

Step 3: The object’s acceleration in the radial direction is given by aradial =v2

Tsin θ−mg cos θ=mv2

Step 4: Solve for the tension T.

T=mv2

L+mg cos θ

Therefore, the expression for the tension Tin terms of m,L,v, and θis T=

mv2

L+mg cos θ.

Question 2

Question

A satellite is orbiting Earth at an altitude of 400 km above the surface. If the

radius of Earth is 6371 km and the mass of Earth is 5.972 ×1024 kg, determine:

a) The speed of the satellite. b) The period of the satellite’s orbit.

Solution

a) To find the speed of the satellite, we can use the formula for the centripetal

force in circular motion:

Fc=mv2

r=G·M·m

Where: - Fcis the centripetal force, - mis the mass of the satellite, - vis the

speed of the satellite, - ris the distance from the Earth’s center to the satellite,

-Gis the gravitational constant, - Mis the mass of the Earth.

Since the satellite is in free-fall, the centripetal force is provided by the

gravitational force:

mv2

r=G·M·m

Solving for v, we get:

v=rG·M

Substitute the known values:

v=s6.67 ×10−11 ·5.972 ×1024

6371 ×103+ 400 ×103

v=s3.986 ×1014

6771 ×103

v≈√59002776

v≈7684.9 m/s

b) The period of the satellite’s orbit can be calculated using the formula:

T=2πr

Substitute the known values:

T=2π·(6371 ×103+ 400 ×103)

7684.9

T=2π·6771 ×103

7684.9

T≈42439.6×103

7684.9

T≈5524.2 s

Therefore, the speed of the satellite is approximately 7684.9 m/s and the

period of the satellite’s orbit is approximately 5524.2 seconds.

Question 3

Question

A satellite of mass mis in a circular orbit of radius raround a planet of mass

M. If the gravitational force between the satellite and the planet is the only

force acting on the satellite, prove that the orbital speed of the satellite is given

by v=qGM

Solution

Step 1: The gravitational force between the satellite and the planet provides the

centripetal force needed to keep the satellite in circular motion. The equation

for gravitational force is given by Newton’s Law of Universal Gravitation: F=

GMm

r2, where Gis the gravitational constant.

Step 2: The centripetal force required to keep the satellite in circular motion

is given by F=mv2

r, where vis the speed of the satellite.

Step 3: Equating the gravitational force to the centripetal force gives us:

GMm

r2=mv2

Step 4: Simplifying this equation, we find:

v2=GM

Step 5: Taking the square root of both sides gives us the orbital speed of

the satellite:

v=rGM

Therefore, the orbital speed of the satellite is given by v=qGM

Question 4

Question

A satellite is in a circular orbit around a planet with a period of 6 hours. Given

that the radius of the orbit is 4 ×107m, determine the mass of the planet.

Assume the satellite is in a low Earth orbit.

Solution

Step 1: Write down the relevant formula for the period of a satellite in circular

motion: The period of a satellite in circular orbit is given by T= 2πqr3

GM ,

where Tis the period, ris the radius of the orbit, Gis the universal gravitational

constant, and Mis the mass of the planet.

Step 2: Rearrange the formula to solve for M:

T= 2πrr3

T2= 4π2r3

M=4π2r3

GT 2

Step 3: Plug in the given values and solve for M: Given T= 6 hours = 21600

seconds, r= 4 ×107m, and G= 6.67 ×10−11 N m2/kg2, we have:

M=4π2(4 ×107)3

6.67 ×10−11 ×216002

M=4π2×64 ×1021

6.67 ×10−11 ×466560000

M=256π2×1021

3.10952 ×107

M=2.54 ×1022

3.10952 ×107

M≈8.15 ×1014 kg

Therefore, the mass of the planet is approximately 8.15 ×1014 kg.

Question 5

Question

A small object of mass mis tied to a string and swung in a vertical circle of

radius R. At the top of the circle, the tension in the string is twice the weight

of the object. Calculate the speed of the object at the top of the circle.

Solution

Step 1: Draw a free body diagram of the object at the top of the circle. At the

top of the circle, the forces acting on the object are the tension in the string T,

pointing downwards, and the weight of the object mg, pointing downwards.

Step 2: Write out the equations for the forces acting on the object at the

top of the circle. The net force acting on the object at the top of the circle is

given by: T−mg =mv2

Step 3: Use the given information to express the tension in terms of the

weight of the object. Given that the tension in the string is twice the weight of

the object, we have: T= 2mg

Step 4: Substitute the tension into the net force equation. Substitute T=

2mg into the net force equation to get: 2mg −mg =mv2

Step 5: Solve for the speed of the object. Solving the equation gives: mg =

mv2

v2=Rg

v=√Rg

Therefore, the speed of the object at the top of the circle is √Rg.

Question 6

Question

A satellite is in a circular orbit around a planet at an altitude of 500 km above

the planet’s surface. The satellite has a mass of 1000 kg and the planet has a

mass of 5 ×1024 kg. If the radius of the planet is 6.4×106m, calculate the

speed of the satellite in its orbit.

Solution

Step 1: First, find the acceleration due to gravity at the altitude of the satellite

using the formula for Newton’s law of gravitation:

F=Gm1m2

where - Fis the force of gravity, - Gis the gravitational constant (6.67 ×

10−11 N·m2/kg2), - m1and m2are the masses of the two objects, - ris the

distance between the centers of the two objects.

The acceleration due to gravity at a distance hfrom the planet’s surface is

given by:

a=GM

(R+h)2

where Mis the mass of the planet and Ris the radius of the planet.

Plugging in the values, we get:

a=(6.67 ×10−11 N·m2/kg2)×(5 ×1024 kg)

(6.4×106m + 500 ×103m)2

a8.94 m/s2

Step 2: Next, find the speed of the satellite in the circular orbit using the

centripetal acceleration formula:

Fnet =m·acentripetal

m·a=m·v2

a=v2

v=√r·a

Substitute the values to get:

v=q6.4×106m·8.94 m/s2

v2.11 ×103m/s

Therefore, the speed of the satellite in its orbit is approximately 2.11 ×103

m/s.

Question 7

Question

A satellite is in a circular orbit around a planet with a radius of 1000 km. The

satellite has a mass of 500 kg. If the gravitational force is the only force acting

on the satellite, determine the speed of the satellite.

Solution

Step 1: Write down the formula for the gravitational force acting on the satellite.

The gravitational force acting on the satellite is given by the formula:

F=GmM

where: - Fis the gravitational force, - Gis the gravitational constant (6.67 ×

10−11 Nm2/kg2), - mis the mass of the satellite (500 kg), - Mis the mass of the

planet (assume much larger than the satellite, constant value), - ris the radius

of the circular orbit (1000 km = 106m).

Step 2: Set the gravitational force equal to the centripetal force acting on

the satellite. The centripetal force acting on an object moving in a circular path

is given by:

F=mv2

where: - vis the speed of the satellite.

Setting the gravitational force equal to the centripetal force, we have:

GmM

r2=mv2

Step 3: Solve for the speed of the satellite. Now we can solve for the speed

of the satellite, v:

v=rGM

v=s(6.67 ×10−11 Nm2/kg2)×M

106m

Step 4: Calculate the speed of the satellite. Assuming the mass of the planet

is much larger than the satellite’s mass, we can simplify the equation to:

v=rGM

Given that the mass of the planet is typically on the order of 1024 kg, we

can use M= 1024 kg in the equation to find the speed of the satellite.

Question 8

Question

A satellite is in a circular orbit around Earth at an altitude of 1000 km above

the surface. The mass of Earth is 5.97 ×1024 kg and the radius of Earth is

6.37 ×106m. Find the orbital speed of the satellite.

Solution

Step 1: Find the distance from the center of Earth to the satellite. Given the

altitude of the satellite above the surface is 1000 km, we can add the radius of

Earth (6.37 ×106m) to get the total distance from the center of Earth to the

satellite:

rtotal = 6.37 ×106m + 1000 ×103m

Step 2: Calculate the total distance from the center of Earth to the satellite.

rtotal = 6.37 ×106m + 1000 ×103m=7.37 ×106m

Step 3: Find the gravitational force acting on the satellite. Using Newton’s

law of gravitation:

F=Gm1m2

where G= 6.67 ×10−11 N m2/kg2is the gravitational constant, m1= 5.97 ×

1024 kg is the mass of Earth, m2is the mass of the satellite (assume 1 kg for

simplicity), and r=rtotal = 7.37 ×106m is the distance from the center of

Earth to the satellite.

Step 4: Calculate the gravitational force.

F= 6.67 ×10−11 N m2/kg2×5.97 ×1024 kg ×1 kg

(7.37 ×106m)2

Step 5: Solve for the gravitational force.

F≈9.77 N

Step 6: Equate gravitational force to centripetal force. The gravitational

force provides the centripetal force necessary to keep the satellite in its circular

orbit.

F=mv2

where vis the orbital speed of the satellite.

Step 7: Calculate the orbital speed. Setting the gravitational force equal to

the centripetal force:

mv2

r=F

Step 8: Solve for the orbital speed.

v=rF r

Step 9: Substitute the values to find the orbital speed.

v=s9.77 N ×7.37 ×106m

1 kg

Step 10: Calculate the orbital speed.

v≈p7.21 ×1014 m/s ≈2.68 ×107m/s

Therefore, the orbital speed of the satellite is approximately 2.68 ×107m/s.

Question 9

Question

A small object is placed at the edge of a frictionless horizontal turntable rotating

counterclockwise with an angular velocity of 2 rad/s. The object has a mass of

0.1 kg and is attached to a string of length 0.5 m. What is the tension in the

string when the object makes an angle of 45 degrees with the vertical?

Solution

Step 1: Identify the forces acting on the object when it makes an angle of 45

degrees with the vertical. At this position, the object experiences two forces:

tension in the string and gravity acting vertically downward.

Step 2: Decompose the forces into components. The weight of the object

can be decomposed into two components: - The component perpendicular to the

string is mg cos θ- The component along the direction of the string is mg sin θ

Step 3: Write the equation of motion for the object along the radial direction.

The net force in the radial direction is the centripetal force required to keep the

object in circular motion. T−mg cos θ=mv2

Step 4: Find the velocity of the object. The velocity of the object can be

found using v=rω, where ωis the angular velocity of the turntable. v=

0.5×2 = 1 m/s

Step 5: Substitute the values back into the equation of motion. T−mg cos θ=

mv2

T−(0.1)(9.81)(cos 45◦) = (0.1)(1)2

0.5

T−0.1(9.81) √2

2!= 0.2

T−0.1(9.81) √2

2!= 0.2

Step 6: Solve for the tension in the string. T= 0.2+0.1(9.81) √2

T= 0.2+0.1(9.81) √2

T≈2.89 N

Therefore, the tension in the string when the object makes an angle of 45

degrees with the vertical is approximately 2.89 N.

Question 10

Question

A satellite of mass mis in a circular orbit around a planet of mass Mwith

radius r. The gravitational force between the satellite and the planet provides

the centripetal force necessary to keep the satellite in its circular path. Show

that the period Tof the satellite’s revolution is given by

T= 2πsr3

G(M+m)

where Gis the gravitational constant.

Solution

Step 1: The magnitude of the gravitational force between the satellite and the

planet is given by Newton’s law of universal gravitation:

Fgravity =GmM

where Gis the gravitational constant, mis the mass of the satellite, Mis the

mass of the planet, and ris the distance between their centers.

Step 2: The gravitational force provides the centripetal force required to

keep the satellite in its circular orbit. The centripetal force required for circular

motion is given by:

Fcentripetal =mv2

where vis the velocity of the satellite.

Step 3: Since the gravitational force and the centripetal force are equal, we

have: GmM

r2=mv2

Step 4: Rearranging the above equation, we can express the velocity vin

terms of r:

v=rGM

Step 5: The period Tof the satellite’s revolution is the time it takes for

the satellite to complete one full revolution around the planet. The period Tis

related to the speed of the satellite and the circumference of its circular orbit

2πr by:

T=2πr

Step 6: Substitute the expression for vfrom Step 4 into the equation for T:

T=2πr

qGM

Step 7: Simplify the expression for Tby rationalizing the denominator:

T= 2πrr3

Step 8: The total mass of the system is M+m, so substitute M+mfor M

in the final expression to obtain the desired result:

T= 2πsr3

G(M+m)

Therefore, the period Tof the satellite’s revolution is given by T= 2πqr3

G(M+m).

Question 11

Question

A satellite is in a circular orbit around a planet with a radius of 5000 km. The

satellite completes one full orbit in 8 hours. Calculate the mass of the planet

given that the gravitational constant is 6.67 ×10−11 Nm2/kg2.

Solution

Step 1: First, we need to find the orbital speed of the satellite using the formula

for the circumference of a circle. The circumference of the circular orbit is:

2πr = 2π×5000 km = 10000πkm = 1.0×107m

Step 2: The time taken to complete one full orbit is 8 hours. Convert this

to seconds:

8 hours = 8 ×60 ×60 s = 28800 s

Step 3: The orbital speed of the satellite is the distance traveled divided by

the time taken to travel that distance:

v=1.0×107m

28800 s = 347.22 m/s

Step 4: The centripetal force required to keep an object in circular motion

is provided by the gravitational force, which can be calculated using:

Fc=mv2

Step 5: The gravitational force is given by Newton’s law of gravitation:

Fg=GMm

Step 6: Equating the centripetal force and gravitational force and canceling

out the mass of the satellite gives:

mv2

r=GMm

Step 7: Solving for the mass of the planet gives:

M=v2r

G=(347.22 m/s)2×5.0×106m

6.67 ×10−11 Nm2/kg2

Step 8: Calculating the mass of the planet:

M=(347.22)2×5.0×106

6.67 ×10−11 = 1.65 ×1024 kg

Therefore, the mass of the planet is 1.65 ×1024 kg.

Question 12

Question

A satellite of mass mis in a circular orbit around a planet of mass M. The

satellite’s orbit has a radius rand the satellite travels at a speed v. If the

gravitational force between the satellite and the planet is the only force acting

on the satellite, prove that the satellite’s acceleration is given by a=GM

r2.

Solution

Step 1: The gravitational force between the satellite and the planet provides

the centripetal force required to keep the satellite in its circular orbit. This can

be expressed as: GMm

r2=m·a

Where Gis the gravitational constant, ais the satellite’s acceleration, and vis

its speed.

Step 2: From the definition of acceleration in circular motion, we have the

relation:

a=v2

Step 3: Equating the centripetal force (from Step 1) to the derived expression

for acceleration (from Step 2), we get:

GMm

r2=m·v2

Step 4: Canceling out the mass mfrom both sides, we find:

r2=v2

Step 5: Rearranging the equation gives us the desired result:

a=GM

Therefore, the satellite’s acceleration is given by a=GM

r2.

Question 13

Question

A small satellite of mass mis in a circular orbit around the Earth at a distance

rfrom the Earth’s center. If the satellite is in a geostationary orbit, meaning

it stays above the same point on the Earth’s equator, calculate the satellite’s

orbital speed in terms of rand the gravitational constant G.

Solution

Step 1: The gravitational force between the Earth and the satellite provides the

centripetal force needed to keep the satellite in orbit. Therefore, we have:

GMm

r2=mv2

where Gis the gravitational constant, Mis the mass of the Earth, and vis the

orbital speed of the satellite.

Step 2: Solving for vgives us:

v=rGM

Step 3: Since the satellite is in a geostationary orbit, it takes the same time

to complete one orbit as the Earth does to rotate once on its axis. The time

period of the satellite’s orbit is equal to the time for one complete rotation of

the Earth, or 24 hours.

Step 4: The orbital speed vcan also be expressed in terms of the circumfer-

ence of the satellite’s orbit C= 2πr and the time period Tas:

v=C

Step 5: Substituting v=qGM

rand C= 2πr into the equation above, we

get:

rGM

r=2πr

Step 6: Solving for T, we have:

T=2πr

pGM/r

Step 7: Since T= 24 hours, we can now express the orbital speed vin terms

of rand G:

v=rGM

r=2πr

24 =πr

Therefore, the orbital speed of the satellite in a geostationary orbit is πr

12 .

Question 14

Question

A satellite is in a circular orbit around a planet with a period of 6 hours. If the

radius of the orbit is 2.5×107m, what is the mass of the planet?

Solution

Step 1: First, let’s find the speed of the satellite in its orbit using the formula

for the centripetal acceleration:

acentripetal =v2

The centripetal acceleration is also given by:

acentripetal =4π2r

T2,

where Tis the period of the orbit. Setting the two expressions for acentripetal

equal to each other gives:

r=4π2r

T2.

Step 2: Solving for vin terms of Tand r, we get:

v=2πr

Substitute r= 2.5×107m and T= 6 hours into the equation to find the speed

of the satellite.

Step 3: Next, we can use the gravitational force equation to find the mass

of the planet:

Fgrav =GMm

r2,

where Fgrav is the gravitational force, Gis the gravitational constant, Mis the

mass of the planet, mis the mass of the satellite, and ris the distance from the

center of the planet to the satellite.

Step 4: The gravitational force is also given by:

Fgrav =mv2

where vis the orbital speed of the satellite determined earlier. Equating the

two expressions for Fgrav, we get:

GMm

r2=mv2

Step 5: Solving for the mass of the planet M, we find:

M=v2r

Substitute the values of v,r, and Ginto the equation to determine the mass of

the planet.

Question 15

Question

A satellite of mass morbits a planet of mass Min a circular orbit of radius

r. Determine the minimum kinetic energy the satellite must have in order to

remain in orbit.

Solution

Step 1: The gravitational force between the satellite and the planet provides

the centripetal force necessary to keep the satellite in a circular orbit. Thus, we

have

GMm

r2=mv2

where Gis the gravitational constant, vis the velocity of the satellite, and

Mis the mass of the planet.

Step 2: Rearranging the equation, we find

v=rGM

Step 3: The kinetic energy of the satellite is given by

KE =1

2mv2

Substitute the expression for vinto the equation for kinetic energy:

KE =1

2m rGM

r!2

Step 4: Simplifying, we get

KE =1

2mGM

Step 5: The minimum kinetic energy required for the satellite to remain in

orbit is when the satellite is at its closest distance rto the planet. Therefore,

the minimum kinetic energy is

KEmin =1

2mGM

So, the minimum kinetic energy the satellite must have in order to remain

in orbit is 1

2mGM

Question 16

Question

A satellite is in a circular orbit around a planet of mass M. The satellite’s mass

is mand the radius of its orbit is r. If the speed of the satellite is doubled, what

will be the new radius of its orbit?

Solution

Step 1: Start by writing the formula for the centripetal force required for an

object moving in a circular path:

Fcentripetal =mv2

Step 2: In circular motion, the centripetal force is provided by the gravita-

tional force between the satellite and the planet. So, we have:

GMm

r2=mv2

GMm =rv2

Step 3: Now, let’s consider the scenario where the speed of the satellite is

doubled. The new speed is 2v. The force between the satellite and the planet

must still be equal to GMm/(2r)2, since the mass and the distance between the

satellite and the planet do not change. So, we have:

GMm =m(2v)2

(2r)

GMm =4mv2

GMm =2mv2

Step 4: Equating the two expressions for the force gives:

rv2=2mv2

r2= 2r

r= 2 (Answer)

Therefore, if the speed of the satellite is doubled, the new radius of its orbit

will be doubled as well.

Question 17

Question

A satellite is in a circular orbit around a planet of mass M. If the period of the

satellite is doubled, what will be the effect on the gravitational force between

the satellite and the planet?

Solution

Let Tbe the original period of the satellite, rbe its original radius of orbit,

vbe its original speed, and Fbe the original gravitational force between the

satellite and the planet.

Step 1: Calculate the original speed of the satellite. The centripetal force

required to keep the satellite in circular motion is provided by the gravitational

force:

F=mv2

where mis the mass of the satellite. This force is also equal to the gravitational

force:

F=GMm

Combining these two equations, we get:

GMm

r2=mv2

Solving for v, we find:

v=rGM

Step 2: Calculate the original period Tof the satellite. The period of

circular motion is given by:

T=2πr

v=2πr

pGM/r = 2πrr3

Step 3: Find the original gravitational force F. Substitute v=qGM

rinto

F=mv2

rto get:

F=GMm

Step 4: Determine the new period 2Tof the satellite. If the period is

doubled, the new period 2Tbecomes:

2T= 2 ·2πrr3

GM = 4πrr3

Step 5: Find the new force F′acting between the satellite and the planet.

Using the relation F′=m(v′)2

r, where v′is the new speed, we get:

F′=GMm

Therefore, doubling the period of the satellite does not affect the gravitational

force between the satellite and the planet.

Question 18

Question

A small asteroid of mass 2.0×1011 kg is on a circular orbit around a star.

The radius of the orbit is 3.0×109m and the period of the orbit is 20 years.

Determine the gravitational force exerted by the star on the asteroid.

Solution

Step 1: Find the speed of the asteroid in its orbit using the formula for the

speed of an object in circular motion:

v=2πr

where: v= speed of the asteroid

r= radius of the orbit

T= period of the orbit

Step 2: Substitute the given values into the formula:

v=2π×3.0×109

20 ×365 ×24 ×3600

Step 3: Calculate the speed of the asteroid:

v≈943.96 m/s

Step 4: Find the centripetal acceleration of the asteroid using the formula:

a=v2

where: a= centripetal acceleration

Step 5: Substitute the calculated values into the formula:

a=(943.96)2

3.0×109

Step 6: Calculate the centripetal acceleration of the asteroid:

a≈2.84 ×10−3m/s2

Step 7: Determine the gravitational force exerted by the star on the asteroid

using the formula for gravitational force:

F=ma

where: F= gravitational force

m= mass of the asteroid

a= centripetal acceleration

Step 8: Substitute the given values into the formula:

F= 2.0×1011 ×2.84 ×10−3

Step 9: Calculate the gravitational force exerted by the star on the asteroid:

F≈5.68 ×108N

Therefore, the gravitational force exerted by the star on the asteroid is ap-

proximately 5.68 ×108N.

Question 19

Question

A satellite is in circular orbit around a planet with a period of Tseconds, a

radius of rmeters, and a mass of mkg. If the gravitational force acting on the

satellite is the only force present, determine an expression for the mass mof the

planet in terms of the given quantities.

Solution

Step 1: First, we need to determine the centripetal force acting on the satellite

in its circular orbit. This force is provided by the gravitational force between

the satellite and the planet. The centripetal force required for circular motion

is given by:

Fc=mv2

where mis the mass of the satellite, vis its velocity, and ris the radius of the

orbit.

Step 2: We can relate the velocity of the satellite to the radius of its orbit

and the period of its motion. The velocity of the satellite in a circular orbit is

given by:

v=2πr

where Tis the period of the orbit.

Step 3: Substituting the expression for velocity into the centripetal force

equation, we get:

Fc=m2πr

T2

Step 4: This centripetal force is provided by the gravitational force between

the satellite and the planet:

Fc=Gmpm

where mpis the mass of the planet and Gis the gravitational constant.

Step 5: Setting the expressions for centripetal force equal to each other, we

get:

m2πr

T2

r=Gmpm

Step 6: Simplifying the equation gives us:

m=4π2r3

GT 2

Step 7: Therefore, the mass mpof the planet in terms of the given quantities

is:

mp=4π2r3

GT 2

Question 20

Question

A satellite is in a circular orbit around a planet with a radius of 1.5×107

meters. If the satellite completes one full orbit in 24 hours, what is the mass of

the planet? Assume the satellite’s mass is negligible compared to the planet’s

mass.

Solution

Step 1: First, we need to find the velocity of the satellite in its circular orbit

using the formula for orbital velocity:

v=2πr

where: v= velocity of the satellite, r= radius of the orbit, T= time period of

one orbit.

Plugging in the values given:

v=2π×1.5×107

24 ×3600

v=3π×107

86400

v≈3495 m/s

Step 2: Next, we can find the gravitational force exerted on the satellite by

the planet using the centripetal force equation:

Fgravity =mv2

where: Fgravity = gravitational force on the satellite, m= mass of the satellite,

v= velocity of the satellite, r= radius of the orbit.

Step 3: Since the satellite’s mass is negligible compared to the planet’s mass,

this gravitational force is provided by the planet’s mass:

Fgravity =GMm

where: G= gravitational constant, M= mass of the planet, r= distance

between the center of the planet and the satellite.

Setting the two expressions for gravitational force equal and solving for M:

GMm

r2=mv2

GM =v2r

M=v2r

M=(3495)2×1.5×107

6.67 ×10−11

M=12252025 ×1.5×107

6.67 ×10−11

M≈183780375000

6.67 ×10−11

M≈2.755 ×1018 kg

Therefore, the mass of the planet is approximately 2.755 ×1018 kg.

Question 21

Question

A satellite is in a circular orbit around Earth at an altitude of 600 km above the

surface of the Earth. The satellite completes one orbit in 90 minutes. Calculate

the magnitude of the acceleration of the satellite. Given that the radius of the

Earth is 6400 km and the acceleration due to gravity at the surface of the Earth

is 9.81 m/s2.

Solution

Step 1: Calculate the total distance traveled by the satellite in one complete

orbit. The radius of the satellite’s orbit can be calculated as the sum of the

radius of the Earth and the altitude of the satellite:

r= radius of Earth + altitude of satellite

r= 6400 km + 600 km = 7000 km = 7 ×106m

The circumference of the satellite’s orbit is given by:

s= 2πr

Step 2: Find the speed of the satellite in its orbit. The time taken for one

complete orbit is 90 minutes, which is equal to 5400 seconds. The speed of the

satellite can be calculated as:

v=s

time taken

v=2πr

5400

Step 3: Calculate the acceleration of the satellite. The acceleration of an

object moving in a circle is given by:

a=v2

Substitute the expression for speed we found in Step 2 into the formula for

acceleration:

a=2πr

5400 2

a=(2π)2r

54002

a=4π2×7×106

54002

a≈8.55 m/s2

Therefore, the magnitude of the acceleration of the satellite is approximately

8.55 m/s2.

Question 22

Question

A satellite is in a circular orbit around Earth at an altitude of 500 km above

the surface. Calculate the satellite’s speed and period of revolution. (Radius of

Earth = 6371 km, mass of Earth = 5.972 ×1024 kg, gravitational constant =

6.674 ×10−11 N m2/kg2)

Solution

Step 1: Calculate the total distance from the center of the Earth to the satellite’s

position: The 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 = 6.871 ×106m

Step 2: Calculate the gravitational force on the satellite: The gravitational

force between the satellite and Earth provides the centripetal force necessary to

keep the satellite in orbit:

Fc=G·M·m

where G= 6.674 ×10−11 N m2/kg2(gravitational constant), M= 5.972 ×1024

kg (mass of Earth), r= 6.871 ×106m (radius of orbit), m(mass of satellite) is

canceled out.

Step 3: Equate the gravitational force to the centripetal force:

G·M·m

r2=m·v2

G·M=v2·r

Step 4: Solve for the satellite’s speed:

v=rG·M

v=r6.674 ×10−11 ×5.972 ×1024

6.871 ×106

v≈7691 m/s

Therefore, the speed of the satellite is approximately 7691 m/s.

Step 5: Calculate the period of revolution: The period of the satellite’s orbit

is given by:

T=2πr

Substitute the values of rand vinto the formula:

T=2π×6.871 ×106

7691

T≈5597 s

Therefore, the period of the satellite’s revolution is approximately 5597 sec-

onds or 93.3 minutes.

Question 23

Question

A satellite is in a circular orbit around a planet with a radius of 1000 km. If

the satellite completes one full orbit in 2 hours, what is the mass of the planet

in kilograms? Assume the gravitational constant is 6.67 ×10−11 m3/kg ·s2.

Solution

Step 1: Find the orbital speed of the satellite using the formula for the speed

in a circular orbit:

v=2πr

where: v= orbital speed (m/s), r= 1000 km = 1,000,000 m (radius of orbit),

T= 2 hours = 7200 s (time for one full orbit), 2π≈6.28 is the constant pi.

Step 2: Substitute the values into the equation and calculate:

v=2·6.28 ·1,000,000

7200 ≈872,665 m/s

Step 3: Use the formula for centripetal force to find the mass of the planet:

Fc=mv2

r=Gm1m2

where: Fc= centripetal force, m= mass of the satellite, v= 872,665 m/s

(orbital speed), G= 6.67 ×10−11 m3/kg ·s2(gravitational constant), r=

1,000,000 m (radius of orbit), m1= mass of the planet.

Step 4: Substitute the values into the equation, simplify, and solve for the

mass of the planet:

mv2

r=Gm1m2

m·872,6652=6.67 ×10−11 ·m·m1

1,000,0002

m1=m·872,6652·1,000,0002

6.67 ×10−11 ·1,000,0002

m1=m·872,6652

6.67 ×10−11

Therefore, the mass of the planet is m·872,6652

6.67×10−11 kg.

Question 24

Question

A satellite is in a circular orbit around Earth at an altitude where the accel-

eration due to gravity is g/2, where gis the acceleration due to gravity at the

surface of Earth. If the radius of the orbit is R, what is the period of the

satellite’s orbit?

Solution

Step 1: We start by considering the gravitational force acting on the satellite.

At the satellite’s orbit, the gravitational force is provided by the gravitational

force equation at a distance Rfrom the center of the Earth:

Fgravity =GmM

where Gis the gravitational constant, mis the mass of the satellite, and Mis

the mass of the Earth.

Step 2: The gravitational force acting on the satellite provides the centripetal

force required to keep the satellite in its circular orbit. Therefore, we equate

the gravitational force to the centripetal force:

GmM

R2=mv2

where vis the orbital speed of the satellite.

Step 3: The orbital speed vcan be expressed in terms of the period Tof the

satellite’s orbit and the radius Rof the orbit:

v=2πR

Step 4: Substituting the expression for vinto the equation from Step 2, we

get:

GmM

R2=m(2πR/T )2

Step 5: Simplifying the equation, we find:

R2=4π2R

Step 6: Solving for the period T, we get:

T= 2πrR3

Step 7: Since the acceleration due to gravity at the satellite’s orbit is g/2,

we can express GM in terms of gand R:

GM = (g/2)R2

Step 8: Substitute the expression for GM into the equation for the period

Tfrom Step 6 to find the final expression for the period:

T= 2πsR3

gR2/2= 2πs2R

Therefore, the period of the satellite’s orbit is 2πq2R

Question 25

Question

A particle of mass 0.2 kg moves in a circular path of radius 0.5 m. If the speed

of the particle is given by v(t) = 3tm/s, where tis in seconds, determine the

magnitude of the total acceleration of the particle at t= 2 s.

Solution

Step 1: We first find the acceleration due to the change in speed. Given v(t) = 3t

m/s, the acceleration a(t) is the derivative of the velocity function v(t) with

respect to time t.

a(t) = d

dtv(t) = d

dt(3t) = 3 m/s2

Step 2: Next, we find the acceleration due to the change in direction in

circular motion. The acceleration due to the change in direction in circular

motion is given by ac=v2

r, where vis the speed of the particle and ris the

radius of the circular path. At t= 2 s, the speed of the particle is v(2) = 3(2) = 6

m/s. Therefore, the centripetal acceleration acis

ac=v2

r=62

0.5= 72 m/s2

Step 3: Finally, we find the total acceleration at t= 2 s by combining the

acceleration due to the change in speed and the centripetal acceleration. The

total acceleration atotal is the vector sum of the acceleration due to the change

in speed and the centripetal acceleration.

atotal =pa2+a2

c=p(3)2+ (72)2≈√9 + 5184 ≈√5193 ≈72 m/s2

Therefore, the magnitude of the total acceleration of the particle at t= 2 s is

approximately 72 m/s2.

Question 26

Question

A satellite is in circular orbit around a planet with a period of 6 hours. If the

satellite is 3 times further from the center of the planet than the planet’s surface

radius, what is the acceleration due to gravity experienced by the satellite in

terms of the acceleration due to gravity at the planet’s surface?

Solution

Step 1: Find the orbital radius of the satellite. Let Rbe the radius of the

planet’s surface, and let rbe the distance from the center of the planet to the

satellite. Given that the satellite is 3 times further from the center of the planet

than the planet’s surface radius, we have r= 3R.

Step 2: Find the orbital speed of the satellite. The orbital speed vof the

satellite can be determined from the formula:

v=2πr

where Tis the period of the satellite’s orbit. Substituting r= 3Rand T= 6

hours into the formula, we get:

v=2π·3R

6=πR

Step 3: Find the acceleration due to gravity at the satellite’s position. The

centripetal acceleration acof the satellite in circular motion can be calculated

using the formula:

ac=v2

Substitute v=πR

3and r= 3Rinto the formula to get:

ac=πR

32

3R=π2

Therefore, the acceleration due to gravity experienced by the satellite in

terms of the acceleration due to gravity at the planet’s surface is π2

9g.

Question 27

Question

A satellite of mass mis moving in a circular orbit of radius raround a planet

of mass M. The gravitational force acting on the satellite is the only force in

the system. Prove that the speed of the satellite is given by v=qGM

Solution

Step 1: The centripetal force required to keep an object moving in a circle is

provided by the gravitational force between the satellite and the planet. Let the

gravitational force on the satellite be given by F=GM m

r2. This force provides

the necessary centripetal force for circular motion, so Fcentripetal =mv2

Step 2: Equate the gravitational force to the centripetal force:

GMm

r2=mv2

GMm =mv2r

Step 3: Cancel out the mass of the satellite, m:

GM =v2r

Step 4: Solve for vto find the speed of the satellite:

v=rGM

Therefore, the speed of the satellite in a circular orbit around a planet of

mass Mat a radius ris v=qGM

Question 28

Question

A satellite of mass mis in a circular orbit around a planet of mass M. The

radius of the orbit is r. If the magnitude of the gravitational force between the

satellite and the planet is F, what is the speed of the satellite in terms of G,

M,m, and r?

Solution

Step 1: The gravitational force between the satellite and the planet is given by

the equation:

F=G·M·m

where Gis the universal gravitational constant.

Step 2: In a circular orbit, the centripetal force required to keep the satellite

in its orbit is provided by the gravitational force:

Fcentripetal =m·v2

where vis the speed of the satellite.

Step 3: Setting Fcentripetal equal to F, we have:

m·v2

r=G·M·m

Step 4: Simplifying the equation, we find the speed of the satellite:

v2=G·M

Step 5: Taking the square root of both sides, we get:

v=rG·M

Therefore, the speed of the satellite in terms of G,M,m, and ris qG·M

Question 29

Question

A satellite of mass mis in a circular orbit around a planet of mass M. Given

the radius of the orbit R, find an expression for the satellite’s orbital velocity.

Solution

Step 1: The gravitational force between the satellite and the planet provides

the centripetal force required for the circular motion of the satellite. Therefore,

we have: GMm

R2=mv2

where Gis the gravitational constant, vis the orbital velocity of the satellite,

and Ris the radius of the orbit.

Step 2: From the equation above, we can solve for the orbital velocity v:

v=rGM

Step 3: Therefore, the expression for the orbital velocity of the satellite is

qGM

Question 30

Question

A satellite of mass mis in a circular orbit around a planet of mass Mand radius

r. The satellite moves with a constant speed of v. Determine the period of the

satellite’s motion in terms of m,M,r, and fundamental constants of nature.

Solution

Step 1: Calculate the gravitational 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:

Fgravity =GMm

Step 2: Equate the gravitational force and the centripetal force. To keep the

satellite in a circular orbit, the gravitational force must equal the centripetal

force: GMm

r2=mv2

Step 3: Solve for the period of the satellite’s motion. The period Tof the

satellite’s motion can be found by rearranging the equation for centripetal force:

T=2πr

Step 4: Substitute the expression for velocity in terms of gravitational force.

From step 2, we have:

v=rGM

Step 5: Substitute the expression for velocity into the equation for period.

Substitute vinto the expression for T:

T=2πr

qGM

Step 6: Simplify the expression for the period. Simplify the expression for

Tto obtain a final expression for the period of the satellite’s motion:

T= 2πrr3

T≈5524.2 s

Therefore, the speed of the satellite is approximately 7684.9 m/s and the

period of the satellite’s orbit is approximately 5524.2 seconds.

Question 3

Question

A satellite of mass mis in a circular orbit of radius raround a planet of mass

M. If the gravitational force between the satellite and the planet is the only

force acting on the satellite, prove that the orbital speed of the satellite is given

by v=qGM

Solution

Step 1: The gravitational force between the satellite and the planet provides the

centripetal force needed to keep the satellite in circular motion. The equation

for gravitational force is given by Newton’s Law of Universal Gravitation: F=

GMm

r2, where Gis the gravitational constant.

Step 2: The centripetal force required to keep the satellite in circular motion

is given by F=mv2

r, where vis the speed of the satellite.

Step 3: Equating the gravitational force to the centripetal force gives us:

GMm

r2=mv2

Step 4: Simplifying this equation, we find:

v2=GM

Step 5: Taking the square root of both sides gives us the orbital speed of

the satellite:

v=rGM

Therefore, the orbital speed of the satellite is given by v=qGM

Question 4

Question

A satellite is in a circular orbit around a planet with a period of 6 hours. Given

that the radius of the orbit is 4 ×107m, determine the mass of the planet.

Assume the satellite is in a low Earth orbit.

Solution

Step 1: Write down the relevant formula for the period of a satellite in circular

motion: The period of a satellite in circular orbit is given by T= 2πqr3

GM ,

where Tis the period, ris the radius of the orbit, Gis the universal gravitational

constant, and Mis the mass of the planet.

Step 2: Rearrange the formula to solve for M:

T= 2πrr3

T2= 4π2r3

M=4π2r3

GT 2

Step 3: Plug in the given values and solve for M: Given T= 6 hours = 21600

seconds, r= 4 ×107m, and G= 6.67 ×10−11 N m2/kg2, we have:

M=4π2(4 ×107)3

6.67 ×10−11 ×216002

M=4π2×64 ×1021

6.67 ×10−11 ×466560000

M=256π2×1021

3.10952 ×107

M=2.54 ×1022

3.10952 ×107

M≈8.15 ×1014 kg

Therefore, the mass of the planet is approximately 8.15 ×1014 kg.

Question 5

Question

A small object of mass mis tied to a string and swung in a vertical circle of

radius R. At the top of the circle, the tension in the string is twice the weight

of the object. Calculate the speed of the object at the top of the circle.

Solution

Step 1: Draw a free body diagram of the object at the top of the circle. At the

top of the circle, the forces acting on the object are the tension in the string T,

pointing downwards, and the weight of the object mg, pointing downwards.

Step 2: Write out the equations for the forces acting on the object at the

top of the circle. The net force acting on the object at the top of the circle is

given by: T−mg =mv2

Step 3: Use the given information to express the tension in terms of the

weight of the object. Given that the tension in the string is twice the weight of

the object, we have: T= 2mg

Step 4: Substitute the tension into the net force equation. Substitute T=

2mg into the net force equation to get: 2mg −mg =mv2

Step 5: Solve for the speed of the object. Solving the equation gives: mg =

mv2

v2=Rg

v=√Rg

Therefore, the speed of the object at the top of the circle is √Rg.

Question 6

Question

A satellite is in a circular orbit around a planet at an altitude of 500 km above

the planet’s surface. The satellite has a mass of 1000 kg and the planet has a

mass of 5 ×1024 kg. If the radius of the planet is 6.4×106m, calculate the

speed of the satellite in its orbit.

Solution

Step 1: First, find the acceleration due to gravity at the altitude of the satellite

using the formula for Newton’s law of gravitation:

F=Gm1m2

where - Fis the force of gravity, - Gis the gravitational constant (6.67 ×

10−11 N·m2/kg2), - m1and m2are the masses of the two objects, - ris the

distance between the centers of the two objects.

The acceleration due to gravity at a distance hfrom the planet’s surface is

given by:

a=GM

(R+h)2

where Mis the mass of the planet and Ris the radius of the planet.

Plugging in the values, we get:

a=(6.67 ×10−11 N·m2/kg2)×(5 ×1024 kg)

(6.4×106m + 500 ×103m)2

a8.94 m/s2

Step 2: Next, find the speed of the satellite in the circular orbit using the

centripetal acceleration formula:

Fnet =m·acentripetal

m·a=m·v2

a=v2

v=√r·a

Substitute the values to get:

v=q6.4×106m·8.94 m/s2

v2.11 ×103m/s

Therefore, the speed of the satellite in its orbit is approximately 2.11 ×103

m/s.

Question 7

Question

A satellite is in a circular orbit around a planet with a radius of 1000 km. The

satellite has a mass of 500 kg. If the gravitational force is the only force acting

on the satellite, determine the speed of the satellite.

Solution

Step 1: Write down the formula for the gravitational force acting on the satellite.

The gravitational force acting on the satellite is given by the formula:

F=GmM

where: - Fis the gravitational force, - Gis the gravitational constant (6.67 ×

10−11 Nm2/kg2), - mis the mass of the satellite (500 kg), - Mis the mass of the

planet (assume much larger than the satellite, constant value), - ris the radius

of the circular orbit (1000 km = 106m).

Step 2: Set the gravitational force equal to the centripetal force acting on

the satellite. The centripetal force acting on an object moving in a circular path

is given by:

F=mv2

where: - vis the speed of the satellite.

Setting the gravitational force equal to the centripetal force, we have:

GmM

r2=mv2

Step 3: Solve for the speed of the satellite. Now we can solve for the speed

of the satellite, v:

v=rGM

v=s(6.67 ×10−11 Nm2/kg2)×M

106m

Step 4: Calculate the speed of the satellite. Assuming the mass of the planet

is much larger than the satellite’s mass, we can simplify the equation to:

v=rGM

Given that the mass of the planet is typically on the order of 1024 kg, we

can use M= 1024 kg in the equation to find the speed of the satellite.

Question 8

Question

A satellite is in a circular orbit around Earth at an altitude of 1000 km above

the surface. The mass of Earth is 5.97 ×1024 kg and the radius of Earth is

6.37 ×106m. Find the orbital speed of the satellite.

Solution

Step 1: Find the distance from the center of Earth to the satellite. Given the

altitude of the satellite above the surface is 1000 km, we can add the radius of

Earth (6.37 ×106m) to get the total distance from the center of Earth to the

satellite:

rtotal = 6.37 ×106m + 1000 ×103m

Step 2: Calculate the total distance from the center of Earth to the satellite.

rtotal = 6.37 ×106m + 1000 ×103m=7.37 ×106m

Step 3: Find the gravitational force acting on the satellite. Using Newton’s

law of gravitation:

F=Gm1m2

where G= 6.67 ×10−11 N m2/kg2is the gravitational constant, m1= 5.97 ×

1024 kg is the mass of Earth, m2is the mass of the satellite (assume 1 kg for

simplicity), and r=rtotal = 7.37 ×106m is the distance from the center of

Earth to the satellite.

Step 4: Calculate the gravitational force.

F= 6.67 ×10−11 N m2/kg2×5.97 ×1024 kg ×1 kg

(7.37 ×106m)2

Step 5: Solve for the gravitational force.

F≈9.77 N

Step 6: Equate gravitational force to centripetal force. The gravitational

force provides the centripetal force necessary to keep the satellite in its circular

orbit.

F=mv2

where vis the orbital speed of the satellite.

Step 7: Calculate the orbital speed. Setting the gravitational force equal to

the centripetal force:

mv2

r=F

Step 8: Solve for the orbital speed.

v=rF r

Step 9: Substitute the values to find the orbital speed.

v=s9.77 N ×7.37 ×106m

1 kg

Step 10: Calculate the orbital speed.

v≈p7.21 ×1014 m/s ≈2.68 ×107m/s

Therefore, the orbital speed of the satellite is approximately 2.68 ×107m/s.

Question 9

Question

A small object is placed at the edge of a frictionless horizontal turntable rotating

counterclockwise with an angular velocity of 2 rad/s. The object has a mass of

0.1 kg and is attached to a string of length 0.5 m. What is the tension in the

string when the object makes an angle of 45 degrees with the vertical?