Question

Find the total energy and binding energy of an artificial satellite of mass $1000kg$ orbiting at height of $1600km$ above the earth's surface. Given $G=6.67\times {10}^{-11}N{m}^2/k{g}^2,R=6400km,M=6\times {10}^{24}kg$

Answer 1

Given

$m=1000$kg $\to$ mass of satellite

$h=1600$km

$G=6.67\times {10}^{-11}\frac{N{m}^2}{k{g}^2}$

$R=6400$km

$M=6\times {10}^{24}$kg

To find: Total energy$=?$

Binding energy$=?$

Total energy(T.E)$=-\frac{GMm}{2(R+h)}$

$=-\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times 1000}{2\times (1600+6400)\times {10}^3}$

$=-\frac{6.67\times {10}^{-11}\times {10}^{24}\times 6}{2\times 8000}$

$=-\frac{6.67\times 6}{16}\times {10}^{10}$

$=-2.50125\times {10}^{10}$J

Binding energy$=-TE$

$=2.50125\times {10}^{10}$J.