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 5.82×10⁶ 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 m_b be the mass of planet Big Alpha and m_o 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 m_b m_o / R² = m_o a
Simplify to obtain
m_b = a R² / G = 3 (5.82×10⁶)² / (6.674×10^-11) = 1.52×10²⁴ 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 m_m be the mass of planet Manta and m_o 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 m_m m_o / R² = m_o a
Simplify to obtain
R² = G m_m / a
R = √ ( G m_m / a ) = √ [ ( 6.674×10^-11)(2.3 × 10²³) / 7 ] = 1.48 × 10⁶ m

Problem 3:
A 1500 kg satellite orbits the Earth at an altitude of 2.5×10⁶ 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 F_c and universal F_u forces have to be equal in magnitude.
F_c = m v² / R , v orbital speed of satellite, m mass of the satellite and R orbital radius
R = Radius of Earth + altitutde = 6.4×10⁶ m + 2.5×10⁶ m = 6.9×10⁶ m
F_u = G M m / R² , M mass of planet Earth
G M m / R² = m v² / R
Simplify to obtain
v = √ (G M / R) = √ [ (6.67×10^-11)(5.96×10²⁴)/(6.9×10⁶) ] = 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 = 2π×6.371×10⁶ / 7590 = 5274 s
c) The kinetic energy E_k of the satellite is given by
E_k = (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 g_m 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 g_m = G M m / R_m² , on the surface of Mars
Simplify to obtain
g_m = G M / R_m²
where M (= 6.39 × 10²³kg) is the mass of Mars, R_m (= 3.39 × 10⁶m) is radius of Mars.
On the surface of Mars
F = m g_m and F = 20 N
m = F / g_m = 20 / g_m
On the surface of the Earth
F_e = g m = 9.8 × F / g_m = 9.8 × 20 / (G M / R_m²) = 9.8×20 × R_m² / (G M)
= 9.8×20 × (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 E_k1 and E_k2 be the kinetic energies of the satellite and v₁ and v₂ the orbital speeds in the first and the second orbits respectively.
E_k1 = (1/2) m v₁² = (1/2) 500 (2πR₁ / T₁)²
E_k2 = (1/2) m v₂² = (1/2) 500 (2πR₂ / T₂)²
E_k2 - E_k1 = 1000 π² [(R₂ / T₂)² - (R₁ / T₁)² ] = 1000 π² [ (10×10⁶ / (8.34×60×60))² - (24×10⁶ / (31×60×60))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 = [ 5.96×10²⁴ × 6.67×10^-11(24×60×60)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 E_k of the satellite is given by
E_k = (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 E_p = - 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 = 6.67×10^-11 × 4.2 × 10²³ × 500 / 4.8 × 10⁹ = 2,919 km
b) Kinetic energy E_k is given by
E_k = (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
E_k = (1/2) m v² = (1/2) G M m / R = (1/2) 4.8 × 10⁹ = 2.4 × 10⁹ J
c) Totale energy E_t is given by
E_t = E_p + E_k = - 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 (5×10⁶)³ / (6.67×10^-11×7.35×10²²)]^1/2 = 8.81 hours

Problem 9:
What is the acceleration on the surface of the Moon?
Solution to Problem 9:
The acceleration g_m 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.
g_m 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 g_m
g_m = G M / R² = 6.67×10^-11×7.35×10²² / 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 = ( 6.67×10^-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

More References and Links

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