Initially a satellite of $100\ \mathrm{kg}$ is in a circular orbit of radius $1.5R_E$. This satellite can be moved to a circular orbit of radius $3R_E$ by supplying $\alpha imes10^6\ \mathrm{J}$ of energy. The value of $\alpha$ is _____.

(Take Radius of Earth $R_E=6 imes10^6\ \mathrm{m}$ and $𝑔=10\ \mathrm{m/s^2}$)

Question

Initially a satellite of $100\ \mathrm{kg}$ is in a circular orbit of radius $1.5R_E$ . This satellite can be moved to a circular orbit of radius $3R_E$ by supplying $\alpha\times10^6\ \mathrm{J}$ of energy. The value of $\alpha$ is _____.

(Take Radius of Earth $R_E=6\times10^6\ \mathrm{m}$ and $𝑔=10\ \mathrm{m/s^2}$ )

(Take Radius of Earth $R_E=6	imes10^6\ \mathrm{m}$ and $𝑔=10\ \mathrm{m/s^2}$)

Accepted Answer

B. 1. The **total mechanical energy** of a satellite of mass $m$ moving in a circular orbit of radius $r$ is

$$
E=K+U,
$$

where

$$
U=-\frac{GMm}{r}.
$$

2. For a circular orbit, the gravitational force provides the centripetal force:

$$
\begin{aligned}
\frac{mv^2}{r}
&=\frac{GMm}{r^2}.
\end{aligned}
$$

Hence,

$$
\begin{aligned}
\frac{1}{2}mv^2
&=\frac{GMm}{2r}.
\end{aligned}
$$

3. Therefore, the total energy is

$$
\begin{aligned}
E
&=\frac{GMm}{2r}-\frac{GMm}{r} \
&=-\frac{GMm}{2r}.
\end{aligned}
$$

4. Using

$$
g=\frac{GM}{R_E^2},
$$

we get

$$
GM=gR_E^2.
$$

Hence,

$$
E
=
-\frac{mgR_E^2}{2r}.
$$

5. The given values are

$$
m=100\ \mathrm{kg},\qquad
r_1=1.5R_E,\qquad
r_2=3R_E.
$$

The energy required is

$$
\Delta E=E_2-E_1.
$$

6. Substituting the energies,

$$
\begin{aligned}
\Delta E
&=-\frac{mgR_E^2}{2r_2}
-\left(-\frac{mgR_E^2}{2r_1}ight) \
&=\frac{mgR_E^2}{2}
\left(
\frac{1}{r_1}
-
\frac{1}{r_2}
ight).
\end{aligned}
$$

7. Replacing the radii,

$$
\begin{aligned}
\Delta E
&=\frac{mgR_E^2}{2}
\left(
\frac{1}{1.5R_E}
-
\frac{1}{3R_E}
ight) \
&=\frac{mgR_E}{2}
\left(
\frac{2}{3}
-
\frac{1}{3}
ight) \
&=\frac{mgR_E}{6}.
\end{aligned}
$$

8. Using

$$
m=100,\qquad
g=10,\qquad
R_E=6	imes10^6\ \mathrm{m},
$$

we obtain

$$
\begin{aligned}
\Delta E
&=\frac{100	imes10	imes6	imes10^6}{6} \
&=1000	imes10^6\ \mathrm{J}.
\end{aligned}
$$

9. Since

$$
\Delta E=\alpha	imes10^6\ \mathrm{J},
$$

it follows that

$$
\boxed{\alpha=1000}.
$$

10. Therefore, **Option B** is correct.