Two identical satellites are at height \(R\) and \(7R\) from earth surface, the ratio of their kinetic energies will be :
Kinetic energy of a satellite is \(K = \frac{GMm}{2r}\). Here \(r_1 = R + R = 2R\) and \(r_2 = R + 7R = 8R\). The ratio \(frac{K_1}{K_2} = \frac{r_2}{r_1} = \frac{8R}{2R} = 4\).
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 \(h = \frac{R k^2}{1-k^2}\).