Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, `T_(A)//T_(B)`, is :

Options

A. `(1)/(2)`
B. 1
C. 2
D. `sqrt((1)/(2))`

Explanation

To solve the problem of finding the ratio of the kinetic energies of two satellites A and B, we can follow these steps:

### Step 1: Understand the Forces Acting on the Satellites
In a circular orbit, the gravitational force provides the necessary centripetal force for the satellite's motion. The gravitational force \( F_g \) acting on a satellite of mass \( m \) at a distance \( r \) from the center of the Earth (mass \( M \)) is given by:

\[ F_g = \frac{GMm}{r^2} \]

Where \( G \) is the universal gravitational constant.

### Step 2: Set Up the Centripetal Force Equation
The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:

\[ F_c = \frac{mv^2}{r} \]

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

### Step 3: Equate Gravitational Force and Centripetal Force
For satellite A (mass \( m \) at radius \( R \)):
\[ \frac{GMm}{R^2} = \frac{mv_A^2}{R} \]

Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{R^2} = \frac{v_A^2}{R} \]

Rearranging gives:
\[ v_A^2 = \frac{GM}{R} \]

### Step 4: Calculate Kinetic Energy of Satellite A
The kinetic energy \( T_A \) of satellite A is given by:
\[ T_A = \frac{1}{2} mv_A^2 = \frac{1}{2} m \left(\frac{GM}{R}\right) = \frac{GMm}{2R} \]

### Step 5: Repeat for Satellite B
For satellite B (mass \( 2m \) at radius \( 2R \)):
\[ \frac{GM(2m)}{(2R)^2} = \frac{(2m)v_B^2}{2R} \]

Cancelling \( 2m \) from both sides:
\[ \frac{GM}{4R^2} = \frac{v_B^2}{2R} \]

Rearranging gives:
\[ v_B^2 = \frac{GM}{2R} \]

### Step 6: Calculate Kinetic Energy of Satellite B
The kinetic energy \( T_B \) of satellite B is given by:
\[ T_B = \frac{1}{2} (2m) v_B^2 = \frac{1}{2} (2m) \left(\frac{GM}{2R}\right) = \frac{GMm}{2R} \]

### Step 7: Find the Ratio of Kinetic Energies
Now, we can find the ratio of the kinetic energies \( \frac{T_A}{T_B} \):
\[ \frac{T_A}{T_B} = \frac{\frac{GMm}{2R}}{\frac{GMm}{2R}} = 1 \]

### Final Answer
Thus, the ratio of the kinetic energies of satellites A and B is:
\[ \frac{T_A}{T_B} = 1 \]