Question

an artificial satellite is moving around earth in a circular orbit equal to one fourth the escape speed of a body , from the surface of earth .the hgt of satellite above earth's surface is (R = radius of earth) ans: 7R why?

Answer 1

Escape velocity = $\sqrt{2gR}$

1/4th escape velocity = $\sqrt{2gR} /4$

=$\sqrt{gR/8}$

= $\sqrt{GM/8R}$

orbital speed = $\sqrt{GM/r}$

so r = 8R

Answer 2

The height of the satellite is 8R

Given:  

Escape velocity of the satellite $=\frac{1}{4} t h$

Solution:  

The escape velocity is given by the formula given below:

$V e=\sqrt{2 g R}$

Given is 1/4th of the escape velocity.

So, we now have,

$V e=\frac{1}{4} \times(\sqrt{2 g R})$

$V e=\sqrt{\frac{g R}{8}}$

The above formula is written as,

$V e=\sqrt{\frac{G M}{8 R}}$

The orbital speed of the orbit of Earth is given by orbit speed is given below

$\Rightarrow \sqrt{\frac{G M}{r}}$

On comparing both the equation, we get,

$r = 8R$