Gravitational Mechanics!

Classical Mechanics Level 4

Consider a hypothetical planet of mass M M and radius R R in deep space such that there is no interaction of any other celestial body on any object nearby it. Now imagine, an object of mass m m taken up to the height of 2 R 2R from the planet's center and then released. Find the time taken by it to hit the surface.

If the time taken can be expressed as: t = R a b ( π c + d ) e G M \large t = \frac{R^{\frac{a}{b}}(\pi^{c} +d )}{e\sqrt{GM}} where a a , b b , c c , d d , and e e are positive integers, find the value of a + b + c + d + e a+b+c+d+e .

Details And Assumptions

  • Assume the planet has no atmosphere.
  • The object is not given any initial energy, except the potential energy it gains.


The answer is 10.

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Kunal Gupta by Kunal Gupta , 23, India

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1 solution

Rui-Xian Siew
Nov 3, 2017

By the law of conservation of energy, the gain of kinetic energy equals to the difference of potential energy. Thus,

1 2 m v 2 = G M m r − G M m 2 R ∴ v = − G M R ( 2 R r − 1 ) \frac { 1 }{ 2 } m{ v }^{ 2 }=\frac { GMm }{ r } -\frac { GMm }{ 2R } \\ \therefore v=-\sqrt { \frac { GM }{ R } (\frac { 2R }{ r } -1) }

Let u = 2 R r − 1 u=\sqrt { \frac { 2R }{ r } -1 }

⇒ d u d r = − R u r 2 = − ( u 2 + 1 ) 2 4 R u ⇒ v = d r d t = d r d u d u d t = − 4 R u ( u 2 + 1 ) 2 d u d t ⇒ d u ( u 2 + 1 ) 2 = G M R − 3 2 4 d t \Rightarrow \frac { du }{ dr } =\frac { -R }{ u{ r }^{ 2 } } =\frac { -{ ({ u }^{ 2 }+1) }^{ 2 } }{ 4Ru } \\ \Rightarrow v=\frac { dr }{ dt } =\frac { dr }{ du } \frac { du }{ dt } =\frac { -4Ru }{ { ({ u }^{ 2 }+1) }^{ 2 } } \frac { du }{ dt } \\ \Rightarrow \frac { du }{ { ({ u }^{ 2 }+1) }^{ 2 } } =\frac { \sqrt { GM } { R }^{ -\frac { 3 }{ 2 } } }{ 4 } dt

Since u = 0 u=0 when r = 2 R r=2R , and u = 1 u=1 when r = R r=R , hence

∫ 0 1 d u ( u 2 + 1 ) 2 = ∫ 0 t G M R − 3 2 4 d t ⇒ G M R − 3 2 4 t = ∫ 0 1 d u ( u 2 + 1 ) 2 \int _{ 0 }^{ 1 }{ \frac { du }{ { ({ u }^{ 2 }+1) }^{ 2 } } } =\int _{ 0 }^{ t }{ \frac { \sqrt { GM } { R }^{ -\frac { 3 }{ 2 } } }{ 4 } dt } \\ \Rightarrow \frac { \sqrt { GM } { R }^{ -\frac { 3 }{ 2 } } }{ 4 } t=\int _{ 0 }^{ 1 }{ \frac { du }{ { ({ u }^{ 2 }+1) }^{ 2 } } } ∫ 0 1 d u ( u 2 + 1 ) 2 = ∫ 0 1 d u ( u 2 + 1 ) − ∫ 0 1 u 2 ( u 2 + 1 ) 2 d u = π 4 + u 2 ( u 2 + 1 ) u = 0 u = 1 − 1 2 ∫ 0 1 d u u 2 + 1 = π 4 + 1 4 − π 8 = π + 2 8 \int _{ 0 }^{ 1 }{ \frac { du }{ { ({ u }^{ 2 }+1) }^{ 2 } } } =\int _{ 0 }^{ 1 }{ \frac { du }{ { ({ u }^{ 2 }+1) } } } -\int _{ 0 }^{ 1 }{ \frac { { u }^{ 2 } }{ { ({ u }^{ 2 }+1) }^{ 2 } } du } \\ =\frac { \pi }{ 4 } +\frac { u }{ 2({ u }^{ 2 }+1) } _{ u=0 }^{ u=1 }-\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { du }{ { { u }^{ 2 }+1 } } } \\ =\frac { \pi }{ 4 } +\frac { 1 }{ 4 } -\frac { \pi }{ 8 } \\ =\frac { \pi +2 }{ 8 }

∴ t = ( π + 2 ) R 3 2 2 G M ∴ a + b + c + d + e = 3 + 2 + 1 + 2 + 2 = 10 \therefore \quad t=\frac { (\pi +2){ R }^{ \frac { 3 }{ 2 } } }{ 2\sqrt { GM } } \\ \therefore \quad a+b+c+d+e=3+2+1+2+2=\boxed { 10 }

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Nice solution !(+1)

Rishu Jaar - 3 years, 7 months ago

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