Question 67: Gravitation

Question

Initially a satellite of 100 kg is in a circular orbit of radius $1.5 \mathrm{R}_{\mathrm{E}}$. This satellite can be moved to a circular orbit of radius $3 R_E$ by supplying $\alpha 	imes 10^6 \mathrm{~J}$ of energy The value of $\alpha$ is $\_\_\_\_$ .
(Take Radius of Earth $R_E=6 	imes 10^6 \mathrm{~m}$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

The total energy ( $E$ ) of a satellite of mass $m$ orbiting a planet of mass $M$ at a radius $r$ is the sum of its Kinetic Energy (K) and Potential Energy (U).
Potential Energy is, $U=-\frac{G M m}{r}$
For a circular orbit, the centripetal force is provided by gravity:
$$ \frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{GMm}}{\mathrm{r}^2} \Rightarrow \frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{GMm}}{2 \mathrm{r}} $$
So, the total energy is,
$$ \mathrm{E}=\mathrm{K}+\mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{r}}-\frac{\mathrm{GMm}}{\mathrm{r}}=-\frac{\mathrm{GMm}}{2 \mathrm{r}} $$
We also know that at the Earth's surface, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{E}}^2}$, so $\mathrm{GM}=\mathrm{gR}_{\mathrm{E}}^2$.
$$ \Rightarrow \mathrm{E}=-\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2 \mathrm{r}} $$
The mass of satellite is, $\mathrm{m}=100 \mathrm{~kg}$
The initial radius is $\mathrm{r}_1=1.5 \mathrm{R}_{\mathrm{E}}$
The final radius is $r_2=3 R_E$
The energy supplied ( $\Delta \mathrm{E}$ ) is the difference between the final total energy and the initial total energy :
$$ \Delta \mathrm{E}=\mathrm{E}_2-\mathrm{E}_1 $$
$$\Rightarrow $$ $\Delta \mathrm{E}=\left(-\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2 \mathrm{r}_2}ight)-\left(-\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2 \mathrm{r}_1}ight)$
$$\Rightarrow $$ $$ \Delta \mathrm{E}=\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2}\left[\frac{1}{\mathrm{r}_1}-\frac{1}{\mathrm{r}_2}ight] $$
Substituting the radii in terms of $\mathrm{R}_{\mathrm{E}}$ :
$$ \Delta \mathrm{E}=\frac{\mathrm{mgR}_{\mathrm{E}}^2}{2}\left[\frac{1}{1.5 \mathrm{R}_{\mathrm{E}}}-\frac{1}{3 \mathrm{R}_{\mathrm{E}}}ight] $$
$$\Rightarrow $$ $\Delta \mathrm{E}=\frac{\mathrm{mgR}_{\mathrm{E}}}{2}\left[\frac{2}{3}-\frac{1}{3}ight]$
$$\Rightarrow $$ $$ \Delta \mathrm{E}=\frac{\mathrm{mgR}_{\mathrm{E}}}{2} 	imes \frac{1}{3}=\frac{\mathrm{mgR}_{\mathrm{E}}}{6} $$
Putting the values, $\mathrm{m}=100, \mathrm{~g}=10, \mathrm{R}_{\mathrm{E}}=6 	imes 10^6$ :
$$ \Delta \mathrm{E}=\frac{100 \cdot 10 \cdot 6 	imes 10^6}{6} $$
$$\Rightarrow $$ $\Delta \mathrm{E}=1000 	imes 10^6 \mathrm{~J}=\alpha 	imes 10^6 \mathrm{~J}$
$$\Rightarrow $$ $$ \alpha 	imes 10^6=1000 	imes 10^6 \Rightarrow \alpha=1000 $$
The value of $\alpha$ required to move the satellite to a higher orbit is 1000 .
Hence, the correct option is (B).

Answer

500

Answer

1000

Answer

100

Answer

150

Gravitation

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