1. Total mechanical energy of a satellite in circular orbit

For a satellite of mass $m$ revolving in a circular orbit of radius $R$ around Earth, the total mechanical energy is

$E = K + U = -\frac{GMm}{2R}$

where $M$ is the mass of Earth.

2. Write data for satellites A and B

• Mass ratio: $m_A : m_B = 4 : 3$
• Orbital radii: $R_A = 3r, \qquad R_B = 4r$

3. Find ratio of total mechanical energies

Using

$E = -\frac{GMm}{2R}$

we get

$E_A : E_B = \left(-\frac{GM m_A}{2R_A}\right) : \left(-\frac{GM m_B}{2R_B}\right)$

Common factors cancel:

$E_A : E_B = \frac{m_A}{R_A} : \frac{m_B}{R_B}$

Substitute values:

$E_A : E_B = \frac{4}{3r} : \frac{3}{4r}$

Cancel $r$:

$E_A : E_B = \frac{4}{3} : \frac{3}{4}$

Now simplify:

$E_A : E_B = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$

Hence,

$E_A : E_B = 16 : 9$

4. Check options

• A: $9:16$ ❌
• B: $16:9$ ✅
• C: $1:1$ ❌
• D: $4:3$ ❌

Therefore, the correct option is B.