If mass of a planet is \(M\) and radius is \(x\), then the work to be done to slowly take a mass \(m\) from surface of planet to a height \(4x\) is :
Initial position \(r_i = x\), final position \(r_f = x + 4x = 5x\). Work done \(W = U_f - U_i = -\frac{GMm}{5x} - \left(-\frac{GMm}{x}\right) = \frac{4GMm}{5x}\).
A body weighs \(900\text{ N}\) on the surface of earth. How much will it weigh at a height double the radius of earth?
Using the formula for acceleration due to gravity at height \(h\): \(g' = g\left(\frac{R}{R+h}\right)^2\). For \(h = 2R\), \(g' = \frac{g}{9}\) . Hence, weight at this height is \(W' = \frac{W}{9} = \frac{900\text{ N}}{9} = 100\text{ N}\).
The escape velocity from the Earth’s surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is
Escape velocity is given by \(v_e = R\sqrt{\frac{8\pi G\rho}{3}}\). Since the mass density \(rho\) is the same, \(v_e\) is directly proportional to radius \(R\). Therefore, \(v_e' = 4v\).
A particle of mass \(m\) is projected with a velocity \(v = k V_e\) (\(k < 1\)) from the surface of the earth. (\(V_e = \text{escape velocity}\)) The maximum height above the surface reached by the particle is
By conservation of mechanical energy: \(-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h}\). Since \(v = k \sqrt{\frac{2GM}{R}}\), we substitute to get \(-\frac{1}{R}(1 - k^2) = -\frac{1}{R+h}\), which yields \(h = \frac{R k^2}{1-k^2}\).