Gravity Problems with Solutions and Explanations

Gravity, problems are presented along with detailed solutions.


Problem 1:

An object is dropped, with no initial velocity, above the surface of planet Big Alpha and falls 13.5 meters in 3 seconds. The radius of planet Big Alpha is 5.82106 meters.

a) What is the acceleration of the falling object?

b) What is the mass of planet Big Alpha?

Solution to Problem 1:

a) Given the distance and the time, we can calculate the acceleration a using the distance formula for the uniform acceleration motion as follows:

d = (1/2) a t 2

a = 2 d / t 2 = 2 13.5 / 3 2 = 3 m/s2

b) Let R be the radius and mb be the mass of planet Big Alpha and mo the mass of the object. The acceleration is due to the universal force of gravity, therefore the universal force of gravity and Newton's second law give

G mb mo / R2 = mo a

Simplify to obtain

mb = a R2 / G = 3 (5.82106)2 / (6.67410-11) = 1.521024 Kg


Problem 2:

An object is dropped, with no initial velocity, near the surface of planet Manta reaches a speed of 21 meters/seconds in 3.0 seconds. Planet Manta has a mass of 2.3 1023 Kg.

a) What is the acceleration acting on the object?

b) What is the radius of planet Manta?

Solution to Problem 2:

a) Given the velocity and the time, we can calculate the acceleration a using the velocity formula of the uniform acceleration motion as follows:

v = a t

a = v / t = 21 / 3 = 7 m/s2

b) Let R be the radius and mm be the mass of planet Manta and mo the mass of the object. The acceleration is due to the universal force of gravity, therefore the universal force of gravity and Newton's second law give

G mm mo / R2 = mo a

Simplify to obtain

R2 = G mm / a

R = √ ( G mm / a ) = √ [ ( 6.67410-11)(2.3 1023) / 7 ] = 1.48 106 m


Problem 3:

A 1500 kg satellite orbits the Earth at an altitude of 2.5106 m.

a) What is the obital speed of the satellite?

b) What is period of the satellite?

c) What is the kinetic energy of the satellite?

Solution to Problem 3:

a) For the satellite to be and stay in orbit, the centripetal Fc and universal Fu forces have to be equal in magnitude.

Fc = m v2 / R , v orbital speed of satellite, m mass of the satellite and R orbital radius

R = Radius of Earth + altitutde = 6.4106 m + 2.5106 m = 6.9106 m

Fu = G M m / R2 , M mass of planet Earth

G M m / R2 = m v2 / R

Simplify to obtain

v = √ (G M / R) = √ [ (6.6710-11)(5.961024)/(6.9106) ] = 7590 m/s

b) The period T is the time it takes the satellite to complete one rotation around the Earth. Hence

T = 2πR / v = 2π6.371106 / 7590 = 5274 s

c) The kinetic energy Ek of the satellite is given by

Ek = (1 / 2) m v2 = (1/2) 1500 75902 = 4.32 1010 J


Problem 4:

The force of gravity that acts on an object on the surface of Mars is 20 N. What force of gravity will act on the same object on the surface of the Earth? (use gravitational field strength g = 9.8 N/Kg on the surface of the Earth).

Solution to Problem 4:

Let the gravitational field strength on Mars be gm and that of Earth be g and m be the mass of the object. The acceleration is due to the universal force of gravity, therefore Newton's second law and the universal force of gravity are equal.

m gm = G M m / Rm2 , on the surface of Mars

Simplify to obtain

gm = G M / Rm2

where M (= 6.39 1023kg) is the mass of Mars, Rm (= 3.39 106m) is radius of Mars.

On the surface of Mars

F = m gm and F = 20 N

m = F / gm = 20 / gm

On the surface of the Earth

Fe = g m = 9.8 F / gm = 9.8 20 / (G M / Rm2) = 9.820 Rm2 / (G M)

= 9.820 (3.39 106)2 / (6.674 10-11 6.39 1023) = 53 N


Problem 5:

A 500 Kg satellite was originally placed into an orbit of radius 24,000 km and a period of 31 hours around planet Barigou.

a) Express the mass of this planet in terms of the Universal constant G, the radius R and the period T.

b) The satellite was then put into its final orbit of radius 10,000km. What was its new period?

c) What is the change in the kinetic energy of the satellite from the first to the second orbits?

Solution to Problem 5:

a) Let M be the mass of the planet and m (=500 Kg) be the mass of the satellite. Satellite orbiting means universal gravitaional force and centripetal forces are equal.

G M m / R2 = m v2 / R , v is the orbital speed of the satellite

Simplify: M = R v2 / G

v = 2πR / T

M = R (2πR / T)2 / G = 4π2 R3 / (G T2)

b) Let T1 and T2 be the period of the satellite at R1 = 24,000,000 and R2 = 10,000,000 m respectively. From the last equation above, we can write

T12 = 4π2 R13 / (M G) and T22 = 4π2 R23 / (M G)

Divide left sides and right sides of the above equations and simplify to obtain

T22 / T12 = R23 / R13

Hence

T2 = √ ( T12 R23 / R13 ) = T1 (R2 / R1 )3/2 = 8.34 hours

c) Let Ek1 and Ek2 be the kinetic energies of the satellite and v1 and v2 the orbital speeds in the first and the second orbits respectively.

Ek1 = (1/2) m v12 = (1/2) 500 (2πR1 / T1)2

Ek2 = (1/2) m v22 = (1/2) 500 (2πR2 / T2)2

Ek2 - Ek1 = 1000 π2 [(R2 / T2)2 - (R1 / T1)2 ] = 1000 π2 [ (10106 / (8.346060))2 - (24106 / (316060))2 ] = 2.30 1012 J


Problem 6:

A 1000 Kg satellite is in synchronous orbit around planet earth. The period of this synchornous orbit matches the rotation of the earth around its axis, assumed to be 24 hours, so that the satellite appears stationary.

a) What is the orbital radius of the satellite?

b) What is the altitude of the satellite?

c) What is the kinetic of the satellite?

Solution to Problem 6:

a) Let M be the mass of the planet and m be the mass of the stellite. Satellite orbiting means universal gravitaional force and centripetal forces are equal

G M m / R2 = m v2 / R , v orbital speed of satellite and R orbital radius

v = 2πR / T

G M m / R2 = m (2πR / T)2 / R

Solve to obtain: R3 = M G T2 / (4π2)

R = [ M G T2 / (4π2) ]1/3 = [ 5.961024 6.6710-11(246060)2 / (4π2) ]1/3 = 42,211 km

b) The radius of the Earth being 6371 km, the altitude h of the satellite is given by

h = 42,211 - 6371 = 35,840 km

c) The kinetic energy Ek of the satellite is given by

Ek = (1/2) m v2 = (1/2) 1000 (2πR / T)2 = (1/2) 1000 (2π 42,211,000 / (24 60 60))2 = 4.7 109 J


Problem 7:

The gravitational potential energy of a 500 kg satellite, orbiting around a planet of mass 4.2 1023, is - 4.8 109 J.

a) What is the orbital radius of this satellite?

b) What is the kinetic energy of this satellite?

c) What is the total energy of this satellite?

d) What is orbital speed of this satellite?

Solution to Problem 7:

a) Use the formula for potetential ebergy Ep = - G M m / R.

- 4.8 109 = - G M m / R

or

G M m / R = 4.8 109

Solve the above for R

R = G M m / 4.8 109 = 6.6710-11 4.2 1023 500 / 4.8 109 = 2,919 km

b) Kinetic energy Ek is given by

Ek = (1/2) m v2 , v orbital speed of satellite

Equality of centripetal and gravitational forces gives

G M m / R2 = m v2 / R

The above equation may be written as: m v2 = G M m / R

and

Ek = (1/2) m v2 = (1/2) G M m / R = (1/2) 4.8 109 = 2.4 109 J

c) Totale energy Et is given by

Et = Ep + Ek = - 4.8 109 + 2.4 109 J = - 2.4 109 J

d) Use kinetic energy (1/2) m v2 found above

(1/2) m v2 = 2.4 109 J

v2 = 2 2.4 109 / m

v = (2 2.4 109 / 500)1/2 = 3,098 m/s


Problem 8:

What is the period of a satellite orbiting the moon at an altitude of 5.0 103 km.

Solution to Problem 8:

Let M be the mass of the moon and m be the mass of the stellite. Satellite orbiting means universal gravitaional force and centripetal forces are equal.

G M m / R2 = m v2 / R , v orbital speed of satellite and R orbital radius

v = 2πR / T , T the period

G M m / R2 = m (2πR / T)2 / R

Solve the above for T to obtain

T = [ 4π2 R3 / G M]1/2

T = [ 4π2 (5106)3 / (6.6710-117.351022)]1/2 = 8.81 hours


Problem 9:

What is the acceleration on the surface of the Moon?

Solution to Problem 9:

The acceleration gm on the surface of the moon is due to the universal force of gravity, therefore Newton's second law and the universal force of gravity are equal.

gm m = G M m / R2 , m mass of any object on the surface of the moon, M mass of the moon and R is the radius of the moon.

Solve for gm

gm = G M / R2 = 6.6710-117.351022 / 1,737,0002 = 1.62 m/s2


Problem 10:

The Hubble Space Telescope orbits the Earth at an altitude of 568 km.

a) What is the orbital speed of the telescope?

b) What is the period of the telescope?

Solution to Problem 10:

a) Let M be the mass of the planet and m be the mass of the telescope. Telescope orbiting means universal gravitaional force and centripetal forces are equal.

G M m / R2 = m v2 / R , v orbital speed of telescope and R its orbital radius

Solve for v

v = ( G M / R)1/2 = ( 6.6710-11 5.96 1024 / (568 103 + 6,400 103) )1/2 = 7553 m/s

b) v = 2πR / T

T = 2πR / T = 2π(568 103 + 6,400 103) / 7553 = 5796 s = 96.6 mn