The ratio of the kinetic energy required to be given to a satellite to escape from the earths gravity to the kinetic energy required to be given to it so that it moves in a circular orbit just above the eart's surface is:

Options

A. `sqrt2`
B. 4
C. `2sqrt2`
D. 2

Explanation

To solve the problem, we need to find the ratio of the kinetic energy required for a satellite to escape Earth's gravity to the kinetic energy required for it to move in a circular orbit just above the Earth's surface.

### Step 1: Calculate the Kinetic Energy Required to Escape Earth's Gravity

The kinetic energy (KE) required for a satellite to escape Earth's gravity can be derived from the concept of escape velocity. The escape velocity (v_e) from the surface of the Earth is given by the formula:

\[ v_e = \sqrt{\frac{2GM}{R}} \]

Where:
- \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\))
- \( M \) is the mass of the Earth (\(5.972 \times 10^{24} \, \text{kg}\))
- \( R \) is the radius of the Earth (\(6.371 \times 10^6 \, \text{m}\))

The kinetic energy required to escape is:

\[ KE_{\text{escape}} = \frac{1}{2} m v_e^2 \]

Substituting for \( v_e \):

\[ KE_{\text{escape}} = \frac{1}{2} m \left(\sqrt{\frac{2GM}{R}}\right)^2 = \frac{1}{2} m \frac{2GM}{R} = \frac{mGM}{R} \]

### Step 2: Calculate the Kinetic Energy Required for Circular Orbit Just Above Earth's Surface

For a satellite to move in a circular orbit just above the Earth's surface, the centripetal force required for circular motion is provided by the gravitational force. The orbital velocity (v_o) for a circular orbit is given by:

\[ v_o = \sqrt{\frac{GM}{R}} \]

The kinetic energy required for this circular motion is:

\[ KE_{\text{orbit}} = \frac{1}{2} m v_o^2 \]

Substituting for \( v_o \):

\[ KE_{\text{orbit}} = \frac{1}{2} m \left(\sqrt{\frac{GM}{R}}\right)^2 = \frac{1}{2} m \frac{GM}{R} \]

### Step 3: Find the Ratio of the Two Kinetic Energies

Now we can find the ratio of the kinetic energy required to escape Earth's gravity to the kinetic energy required for circular orbit:

\[ \text{Ratio} = \frac{KE_{\text{escape}}}{KE_{\text{orbit}}} = \frac{\frac{mGM}{R}}{\frac{1}{2} m \frac{GM}{R}} = \frac{mGM/R}{(1/2)mGM/R} = 2 \]

### Final Answer

The ratio of the kinetic energy required to escape from Earth's gravity to the kinetic energy required to move in a circular orbit just above the Earth's surface is:

\[ \text{Ratio} = 2 \]