1. Kinetic energy of a satellite in circular orbit

For a satellite of mass $m$ in a circular orbit of radius $r$ around Earth,

$\frac{GM_E m}{r^2} = \frac{m v^2}{r}$

So,

$v^2 = \frac{GM_E}{r}$

Hence the kinetic energy is

$T = \frac{1}{2} m v^2 = \frac{1}{2} m \cdot \frac{GM_E}{r} = \frac{GM_E m}{2r}$

Thus,

$T \propto \frac{m}{r}$

2. Kinetic energy of satellite A

Satellite A has mass $m$ and orbit radius $R$.

Therefore,

$T_A = \frac{GM_E m}{2R}$

3. Kinetic energy of satellite B

Satellite B has mass $2m$ and orbit radius $2R$.

Therefore,

$T_B = \frac{GM_E (2m)}{2(2R)} = \frac{GM_E m}{2R}$

4. Find the ratio

$\frac{T_A}{T_B} = \frac{\frac{GM_E m}{2R}}{\frac{GM_E m}{2R}} = 1$

5. Check options

• A: $2$ ❌
• B: $\frac{1}{2}$ ❌
• C: $\sqrt{\frac{1}{2}}$ ❌
• D: $1$ ✅

So the correct answer is D.