The ratio of the binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height h above the earth's surface is (.R = radius of the earth

Question

The ratio of the binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height h above the earth's surface is (.R = radius of the earth (A) (2 (. R + h .)/R) (B)  (R + h/2 R) (C) (R + h/R) (D) (R/R + h)

Tardigrade Admin · Accepted Answer

Binding energy on the surface of the earth E1=(G M m/2 R) Binding energy of revolving satellite E2=(G M m/2 (. R + h .)) ∴ (E1/E2)=((. R + h .)/R)

Q. The ratio of the binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height $h$ above the earth's surface is $(R$ = radius of the earth

$\frac{2 ( R + h )}{R}$

$\frac{R + h}{2 R}$

$\frac{R + h}{R}$

$\frac{R}{R + h}$

Binding energy on the surface of the earth
$E_{1} = \frac{GM m}{2 R}$
Binding energy of revolving satellite
$E_{2} = \frac{GM m}{2 ( R + h )}$
$∴ \frac{E _{1}}{E _{2}} = \frac{( R + h )}{R}$

Q. The ratio of the binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height h above the earth's surface is (R = radius of the earth

R2(R+h)​

2RR+h​

RR+h​

R+hR​