Question 355: Gravitation

Question

The ratio of energy required to raise a satellite of mass $$m$$ to a height $$h$$ above the earth's surface of that required to put it into the orbit at same height is [$$R=$$ radius of the earth]

Accepted Answer

The energy required to raise a satellite of mass m to a height h,
$$E_1=\frac{m g h}{\left(1+\frac{h}{R}ight)} \quad 	ext{... (i)}$$
$$\begin{aligned}
&	ext { The energy required to put it into orbit, }\
&E_2=\frac{1}{2} m v_0^2=\frac{1}{2} m\left(\frac{G M}{r}ight)
\end{aligned}$$
$$\left(\because 	ext { orbital velocity, } v_0=\frac{G M}{r}ight)$$
$$\begin{aligned}
& =\frac{1}{2} m\left(\frac{G M}{R+h}ight)=\frac{1}{2} \frac{m g R^2}{R\left(1+\frac{h}{R}ight)}\left(\because G M=g R^2ight) \
& =\frac{m g R}{2\left(1+\frac{h}{R}ight)}
\end{aligned}$$
$$\begin{aligned}
&	ext { The ratio of } E_1: E_2 	ext { becomes, }\
&\frac{E_1}{E_2}=\frac{m g h}{\left(1+\frac{h}{R}ight)} 	imes \frac{2\left(1+\frac{h}{R}ight)}{m g R}=\frac{2 h}{R}
\end{aligned}$$

Answer

$$\frac{h}{R}$$

Answer

$$\frac{3 h}{R}$$

Answer

$$\frac{4 h}{R}$$

Answer

$$\frac{2 h}{R}$$

Gravitation

The ratio of energy required to raise a satellite of mass $m to a height h above the earth's surface of that required to put it into the orbit at same height is [R=$ radius of the earth]

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