Two artificial satellites A and B of same mass are revolving round the earth at heights `h_1` and `h_2` in circular orbits. If `h_1 lt h_2`, the satellite possessing greater KE is

Options

A. A
B. B
C. Both A & B
D. cannot be determined

Answer

To determine which satellite possesses greater kinetic energy, we can follow these steps:

### Step 1: Understand the relationship between gravitational force and centripetal force
For a satellite in circular orbit, the gravitational force provides the necessary centripetal force to keep the satellite in orbit. The gravitational force acting on a satellite of mass \( M \) at a distance \( R + h \) from the center of the Earth is given by:

\[ F_g = \frac{GMm}{(R + h)^2} \]

where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( R \) is the radius of the Earth.

### Step 2: Set up the equation for centripetal force
The centripetal force required for the satellite to stay in circular motion is given by:

\[ F_c = \frac{mv^2}{R + h} \]

where \( v \) is the orbital velocity of the satellite.

### Step 3: Equate gravitational force and centripetal force
Setting the gravitational force equal to the centripetal force gives us:

\[ \frac{GMm}{(R + h)^2} = \frac{mv^2}{R + h} \]

### Step 4: Solve for the orbital velocity \( v \)
By simplifying the equation, we can cancel \( m \) from both sides (assuming \( m \neq 0 \)) and multiply both sides by \( (R + h) \):

\[ \frac{GM}{(R + h)} = v^2 \]

Thus, we can express the velocity \( v \) as:

\[ v = \sqrt{\frac{GM}{(R + h)}} \]

### Step 5: Calculate the kinetic energy for each satellite
The kinetic energy \( KE \) of a satellite is given by:

\[ KE = \frac{1}{2} mv^2 \]

Substituting the expression for \( v \):

\[ KE = \frac{1}{2} m \left(\frac{GM}{(R + h)}\right) \]

### Step 6: Compare the kinetic energies of satellites A and B
Let satellite A be at height \( h_1 \) and satellite B at height \( h_2 \) (where \( h_1 < h_2 \)). The kinetic energies can be expressed as:

- For satellite A:

\[ KE_A = \frac{1}{2} m \left(\frac{GM}{(R + h_1)}\right) \]

- For satellite B:

\[ KE_B = \frac{1}{2} m \left(\frac{GM}{(R + h_2)}\right) \]

### Step 7: Analyze the relationship between \( KE_A \) and \( KE_B \)
Since \( h_1 < h_2 \), we have:

\[ R + h_1 < R + h_2 \]

This implies:

\[ \frac{1}{R + h_1} > \frac{1}{R + h_2} \]

Thus, we find that:

\[ KE_A > KE_B \]

### Conclusion
The satellite at height \( h_1 \) (satellite A) possesses greater kinetic energy than the satellite at height \( h_2 \) (satellite B).