Question

A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. the total energy of the satellite in terms of g₀, the value of acceleration due to gravity at the earth's surface, is

विकल्प

• ${\displaystyle -\frac{2m{g}_0{R}^2}{R+h}}$

• ${\displaystyle \frac{m{g}_0{R}^2}{2(R+h)}}$

• ${\displaystyle -\frac{m{g}_0{R}^2}{2(R+h)}}$

• ${\displaystyle \frac{2m{g}_0{R}^2}{R+h}}$

Answer 1

${\displaystyle -\frac{m{g}_0{R}^2}{2(R+h)}}$

Explanation:

A satellite's total energy is,

T.E. = ${\displaystyle \frac{GMm}{2(R+h)}}$  ......(i)

∴ Taking the equation (i) and multiplying and dividing it by R²

T.E. = ${\displaystyle \frac{GMm{R}^2}{2(R+h)R^2}}$

∴ T.E. = ${\displaystyle \frac{g_{0}m{R}^2}{2(R+h)}}$  .....${\displaystyle (\because g_0=\frac{GM}{R^2})}$