Question

$E_1$ is energy required to take a satellite up to a height h from surface of earth. $E_2$ is kinetic energy required to perform circular motion at that height. If $E_1=E_2$ then h will be [Radius of earth $6.4\times {10}^3$ km]

• A. $6.4\times {10}^3$ km
• B. $1.6\times {10}^4$ km
• C. $3.2\times {10}^3$ km
• D. $1.6\times {10}^3$ km

Answer 1

The correct option is C $3.2\times {10}^3$ km

$V_0=\sqrt{\frac{G{M}_e}{(R+h)}}$

$KE=E_2=\frac{1}{2}m{v}^2=\frac{G{M}_{e}m}{2(R_e+n)}$

$E_1=\frac{Gm{M}_e}{\left(R_e\right)}=-\frac{G{M}_{e}m}{(R_e+n)}$

since $E_1=E_2$

$\frac{G{M}_{e}m}{2(R_e+n)}-\frac{Gm{M}_e}{R_e}=-\frac{G{M}_{e}m}{(R_e+n)}$

$\frac{3G{M}_{e}m}{2(R_e+n)}=\frac{Gm{M}_e}{R_e}$

$3R_e=2R_e+2h$

$h=R_e/2$