Q159 / 170 · Manipal (MET) 2015 · Physics
The ratio of the energy required to raise a satellite upto a height $h$ above the earth to the kinetic energy of the satellite into the orbit is
- A$h: R$
- B$R: 2 h$
- Correct answerC$2 h: R$
- D$R: h$
Solution
Energy required to raise a satellite upto a height $h$,
$E_1=\Delta U=\frac{m g h}{1+\frac{h}{R}}$ ...(i)
$E_2=$ energy required to put in orbit
$\begin{aligned} & =\frac{1}{2} m v_0^2 \\ & =\frac{1}{2} m\left(\frac{G M}{r}\right) \text { as } v_0=\text { orbital speed } \\ & =\frac{1}{2} m\left(\frac{G M}{R+h}\right)=\frac{1}{2} m\left(\frac{G M}{R^2}\right) \frac{R}{1+\frac{h}{R}}\end{aligned}$ ...(ii)
$\Rightarrow \quad E_2=\frac{m g R}{2\left(1+\frac{h}{R}\right)}$
From Eqs. (i) and (ii), we get
$\frac{E_1}{E_2}=\frac{2 h}{R}$
Hence, the correct answer is option: 'C'.