Gravity Escape velocity

Classical Mechanics Level 4

v e = 2 G M r \large v_e = \sqrt{\frac {2GM}r}

A body is projected upwards with a with a velocity equal to 3 4 \frac 34 the escape velocity of Earth v e v_e . What is the height it reaches above the surface of the Earth? You will get an answer in the form of a R b \dfrac{aR}{b} . Provide the answer as a + b a + b .

For more such problems, try my set Gravity


The answer is 16.

Sravanth C. by Sravanth C. , 21, India

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1 solution

Sravanth C.
Feb 15, 2015

Escape velocity of the earth is v e = 2 G M R { v }_{ e }=\sqrt { \frac { 2GM }{ R } }

Let v be the velocity with which the body is projected.

According to the law of conservation of mechanical energy, we have

1 2 m v 2 − G M m R = − G M m R + h \frac { 1 }{ 2 } m{ v }^{ 2 }-\frac { GMm }{ R } =-\frac { GMm }{ R+h }

or, 1 2 m ( 3 4 v e ) 2 − G M m R = − G M m ( R + h ) 1 2 m ( 3 4 2 G M R ) 2 − G M m R = − G M m ( R + h ) 9 16 G M m R − G M m R = − G M m ( R + h ) 1 R ( 9 16 − 1 ) = − 1 ( R + h ) 7 16 R = 1 R + h 7 R + 7 h = 16 R h = 9 R 7 \frac { 1 }{ 2 } m{ \left( \frac { 3 }{ 4 } { v }_{ e } \right) }^{ 2 }-\frac { GMm }{ R } =-\frac { GMm }{ (R+h) } \\ \frac { 1 }{ 2 } m{ \left( \frac { 3 }{ 4 } \sqrt { \frac { 2GM }{ R } } \right) }^{ 2 }-\frac { GMm }{ R } =-\frac { GMm }{ (R+h) }\\ \frac { 9 }{ 16 } \frac { GMm }{ R } -\frac { GMm }{ R } =-\frac { GMm }{ (R+h) } \\ \frac { 1 }{ R } \left( \frac { 9 }{ 16 } -1 \right) =-\frac { 1 }{ (R+h) }\\\frac { 7 }{ 16R } =\frac { 1 }{ R+h }\\7R+7h=16R \\h=\frac { 9R }{ 7 }

Therefore, 9 + 7 = 16 9+7=\boxed{16}

11 Helpful 0 Interesting 0 Brilliant 0 Confused

Instead of deriving it each time, you can use this formula.

h m a x = v 2 2 g − v 2 R h_{max} = \dfrac{v^2}{2g - \frac{v^2}{R}}

where R is the radius of earth.

It can be seen that when v v is small, the term v 2 R \dfrac{v^2}{R} is neglected.

Vishwak Srinivasan - 5 years, 10 months ago

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