An artifical satellite of mass 'm' is moving in a circular orbit aroundthe earth. The height of the satellite above the surface of the earth is R. Suppose that it stops suddenly in its orbit and falls freely under gravity. With what speed it will strike the surface of the earth?

Options

A. `sqrt(gR)`
B. `2sqrt(gR)`
C. `3sqrt(gR)`
D. `5sqrt(gR)`

Answer

To solve the problem of determining the speed at which an artificial satellite strikes the surface of the Earth after stopping suddenly in its orbit, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Initial and Final States
- The satellite is initially at a height \( R \) above the Earth's surface.
- The radius of the Earth is denoted as \( R \), so the distance from the center of the Earth to the satellite is \( 2R \) (Earth's radius + height of the satellite).

### Step 2: Write the Conservation of Mechanical Energy Equation
- The mechanical energy (sum of potential and kinetic energy) is conserved since the only force acting on the satellite after it stops is gravity.
- The equation can be written as:
\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]

### Step 3: Define Initial and Final Energies
- Initially, the satellite is at rest (stopped), so its initial kinetic energy is:
\[ KE_{initial} = 0 \]
- The initial potential energy \( PE_{initial} \) at a distance \( 2R \) from the center of the Earth is:
\[ PE_{initial} = -\frac{GMm}{2R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the satellite.

- When the satellite strikes the surface of the Earth, its potential energy \( PE_{final} \) at distance \( R \) from the center of the Earth is:
\[ PE_{final} = -\frac{GMm}{R} \]
- Let \( v \) be the final speed of the satellite just before it strikes the surface. The final kinetic energy \( KE_{final} \) is:
\[ KE_{final} = \frac{1}{2}mv^2 \]

### Step 4: Set Up the Energy Conservation Equation
- Plugging in the values into the conservation of energy equation:
\[ 0 - \frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \]

### Step 5: Simplify the Equation
- Rearranging gives:
\[ \frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \]
- Combine the potential energy terms:
\[ \frac{GMm}{2R} + \frac{GMm}{R} = \frac{1}{2}mv^2 \]
- This simplifies to:
\[ \frac{GMm + 2GMm}{2R} = \frac{1}{2}mv^2 \]
- Thus:
\[ \frac{3GMm}{2R} = \frac{1}{2}mv^2 \]

### Step 6: Solve for \( v^2 \)
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{3GM}{2R} = \frac{1}{2}v^2 \]
- Multiply both sides by 2:
\[ \frac{3GM}{R} = v^2 \]

### Step 7: Solve for \( v \)
- Taking the square root gives:
\[ v = \sqrt{\frac{3GM}{R}} \]

### Step 8: Relate \( g \) to \( GM/R^2 \)
- We know that \( g = \frac{GM}{R^2} \), hence:
\[ v = \sqrt{3gR} \]

### Final Answer
The speed at which the satellite will strike the surface of the Earth is:
\[ v = \sqrt{3gR} \]