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Question

Physics Question on Escape Speed

The acceleration due to gravity becomes (g2)\left(\frac{g}{2}\right) where (g = acceleration due to gravity on the surface of the earth) at a height equal to

A

R2\frac{R}{2}

B

2R2R

C

R4\frac{R}{4}

D

4R4R

Answer

R4\frac{R}{4}

Explanation

Solution

The acceleration due to gravity
g=GMR2g=\frac{G M}{R^{2}}
At a height hh above the earth's surface, the acceleration due to gravity is
g=GM(R+h)2g' =\frac{G M}{(R+h)^{2}}
gg=(R+hR)2\frac{g}{g'} =\left(\frac{R+h}{R}\right)^{2}
=(1+hR)2=\left(1+\frac{h}{R}\right)^{2}
gg=(1+hR)2\frac{g'}{g} =\left(1+\frac{h}{R}\right)^{-2}
=(12hR)=\left(1-\frac{2 h}{R}\right)
but gg2 g'\frac{g}{2} (given)
g/2h=12hR\frac{g / 2}{h} =1-\frac{2 h}{R}
2hR=12\frac{2 h}{R} =\frac{1}{2}
h=R4h=\frac{R}{4}