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Solutions to Physics I Gravity and Kepler’s Laws Practice Problems
1.) Titan, the largest moon of Saturn, has a mean orbital radius of 1.22x109 m. The orbital period
of Titan is 15.95 days. Hyperion, another moon of Saturn, orbits at a mean radius of 1.48x109 m.
Use Kepler’s third law of planetary motion to predict the orbital period of Hyperion in days.
π‘Ÿπ‘‡ = 1.22π‘₯109 π‘š
𝑇𝑇 = 15.95 π‘‘π‘Žπ‘¦π‘ 
π‘Ÿπ» = 1.48π‘₯109 π‘š
𝑇𝐻 =?
𝑇𝑇 2
π‘Ÿπ‘‡ 3
( ) =( )
𝑇𝐻
π‘Ÿπ»
3
15.95 π‘‘π‘Žπ‘¦π‘  2
1.22π‘₯109 π‘š
(
) =(
)
𝑇𝐻
1.48π‘₯109 π‘š
15.95 π‘‘π‘Žπ‘¦π‘  2
(
) = 0.8243
𝑇𝐻
254.4 π‘‘π‘Žπ‘¦π‘  2
𝑇𝐻 2
𝑇𝐻 = √
= 0.560
254.4 π‘‘π‘Žπ‘¦π‘  2
0.560
𝑇𝐻 = 21.3 π‘‘π‘Žπ‘¦π‘ 
2.) The mass of Earth is 5.97x1024 kg, the mass of the Moon is 7.35x1022 kg, and the mean
distance of the Moon from the center of Earth is 3.84x105 km. Use these data to calculate the
magnitude of the gravitational force exerted by Earth on the Moon.
π‘šπΈ = 5.97π‘₯1024 π‘˜π‘”
π‘šπ‘€ = 7.35π‘₯1022 π‘˜π‘”
π‘Ÿ = 3.84π‘₯105 π‘˜π‘š = 3.84π‘₯108 π‘š
𝐺 = 6.673π‘₯10βˆ’11 𝑁 βˆ™ π‘š2 β„π‘˜π‘”2
𝐹𝑀𝐸 =?
𝐹𝑀𝐸 = 𝐺
π‘šπΈ π‘šπ‘€
π‘Ÿ2
(5.97π‘₯1024 π‘˜π‘”)(7.35π‘₯1022 π‘˜π‘”)
𝐹𝑀𝐸 = (6.673π‘₯10βˆ’11 𝑁 βˆ™ π‘š2 β„π‘˜π‘”2 ) [
]
(3.84π‘₯108 π‘š)2
4.39π‘₯1047 π‘˜π‘”2
𝐹𝑀𝐸 = (6.673π‘₯10βˆ’11 𝑁 βˆ™ π‘š2 β„π‘˜π‘”2 ) (
)
1.47π‘₯1017 π‘š2
𝐹𝑀𝐸 = (6.673π‘₯10βˆ’11 𝑁 βˆ™ π‘š2 β„π‘˜π‘”2 )(2.98π‘₯1030 π‘˜π‘”2 β„π‘š2 )
𝐹𝑀𝐸 = 1.99π‘₯1020 𝑁
3.) The planet Mercury travels around the Sun with a mean orbital radius of 5.8x1010 m. The
mass of the Sun is 1.99x1030 kg. Use Newton’s version of Kepler’s third law to determine how
long it takes Mercury to orbit the Sun. Give your answer in Earth days.
π‘Ÿπ‘€ = 5.810π‘₯1010 π‘š
π‘šπ‘† = 1.99π‘₯1030 π‘˜π‘”
𝑇𝑀 =?
𝑇𝑀
2
4πœ‹ 2 3
=(
)π‘Ÿ
πΊπ‘šπ‘†
𝑇𝑀 2 = [
𝑇𝑀 2 = [
(6.673π‘₯10βˆ’11
39.5
] (5.810π‘₯1010 π‘š)3
𝑁 βˆ™ π‘š2 β„π‘˜π‘”2 )(1.99π‘₯1030 π‘˜π‘”)
39.5
] (1.96π‘₯1032 π‘š3 )
𝑁 βˆ™ π‘š2 β„π‘˜π‘”
1.33π‘₯1020
𝑇𝑀 2 = (2.96π‘₯10βˆ’19 𝑠 2 β„π‘š3 )(1.96π‘₯1032 π‘š3 )
𝑇𝑀 2 = 5.82π‘₯1013 𝑠 2
𝑇𝑀 = √5.82π‘₯1013 𝑠 2
1 β„Žπ‘œπ‘’π‘Ÿ
1 π‘‘π‘Žπ‘¦
𝑇𝑀 = 7.63π‘₯106 𝑠 (
)(
) = 88.3 π‘‘π‘Žπ‘¦π‘ 
3600 𝑠 24 β„Žπ‘œπ‘’π‘Ÿπ‘ 
4.) Earth has an orbital period of 365 days and its mean distance from the Sun is 1.495x108 km.
The planet Pluto’s mean distance from the Sun is 5.896x109 km. Using Kepler’s third law,
calculate Pluto’s orbital period in Earth days.
𝑇𝐸 = 365 π‘‘π‘Žπ‘¦π‘ 
π‘ŸπΈ = 1.495π‘₯108 π‘˜π‘š
π‘Ÿπ‘ƒ = 5.896π‘₯109 π‘˜π‘š
𝑇𝑃 =?
𝑇𝐸 2
π‘ŸπΈ 3
( ) =( )
𝑇𝑃
π‘Ÿπ‘ƒ
3
365 π‘‘π‘Žπ‘¦π‘  2
1.495π‘₯108 π‘˜π‘š
(
) =(
)
𝑇𝑃
5.896π‘₯109 π‘˜π‘š
365 π‘‘π‘Žπ‘¦π‘  2
(
) = (2.54π‘₯10βˆ’2 )3
𝑇𝑃
1.32π‘₯105 π‘‘π‘Žπ‘¦π‘  2
(
) = 1.63π‘₯10βˆ’5
2
𝑇𝑃
1.32π‘₯105 π‘‘π‘Žπ‘¦π‘  2
𝑇𝑃 = √
1.63π‘₯10βˆ’5
𝑇𝑃 = 9.00π‘₯104 π‘‘π‘Žπ‘¦π‘ 
5.) The planet Venus orbits the Sun with a mean orbital radius of 1.076x1011 m. The mass of the
Sun is 1.99x1030 kg. Using Newton’s version of Kepler’s third law, calculate the orbital period of
Venus.
π‘Ÿπ‘‰ = 1.076π‘₯1011 π‘š
π‘šπ‘† = 1.99π‘₯1030 π‘˜π‘”
𝑇𝑉 =?
𝑇𝑉
2
4πœ‹ 2
=(
)π‘Ÿ 3
πΊπ‘šπ‘† 𝑉
2
𝑇𝑉 = [
(6.673π‘₯10βˆ’11
𝑇𝑉 2 = [
4πœ‹ 2
] (1.076π‘₯1011 π‘š)3
𝑁 βˆ™ π‘š2 β„π‘˜π‘”2 )(1.99π‘₯1030 π‘˜π‘”)
39.5
] (1.25π‘₯1033 π‘š3 )
𝑁 βˆ™ π‘š2 β„π‘˜π‘”
1.33π‘₯1020
𝑇𝑉 2 = (2.97π‘₯10βˆ’19 𝑠 2 β„π‘š3 )(1.25π‘₯1033 π‘š3 )
𝑇𝑉 2 = 3.17π‘₯1014 𝑠 2
𝑇𝑉 = √3.17π‘₯1014 𝑠 2
𝑇𝑉 = 1.93π‘₯107 𝑠