The escape velocity of a sphere of mass m is given by (G= univesal gravitational constant, `M_(e) =` mass of the earth and `R_(e) =` radius of the earth)

Options

A. `sqrt((GM_(e))/(R_(e)))`
B. `sqrt((2GM_(e))/(R_(e)))`
C. `sqrt((2Gm)/(R_(e)))`
D. `(GM_(e))/(R_(e)^(2))`

Answer

To find the escape velocity of a sphere of mass \( m \) from the surface of the Earth, we can follow these steps:

### Step 1: Understand the Concept of Escape Velocity
Escape velocity is the minimum velocity required for an object to break free from the gravitational attraction of a celestial body without any additional propulsion.

### Step 2: Write the Formula for Gravitational Potential Energy
The gravitational potential energy \( U \) of an object of mass \( m \) at a distance \( R_e \) (radius of the Earth) from the center of the Earth (mass \( M_e \)) is given by:
\[ U = -\frac{G M_e m}{R_e} \]
where \( G \) is the universal gravitational constant.

### Step 3: Relate Binding Energy to Kinetic Energy
When the object escapes the gravitational field, the gravitational potential energy will be converted into kinetic energy. The kinetic energy \( K \) of the object when it reaches escape velocity \( V_e \) is given by:
\[ K = \frac{1}{2} m V_e^2 \]

### Step 4: Set Up the Energy Conservation Equation
At the point of escape, the kinetic energy will equal the magnitude of the gravitational potential energy:
\[ \frac{1}{2} m V_e^2 = \frac{G M_e m}{R_e} \]

### Step 5: Simplify the Equation
We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} V_e^2 = \frac{G M_e}{R_e} \]

### Step 6: Solve for Escape Velocity
Now, multiply both sides by 2 to isolate \( V_e^2 \):
\[ V_e^2 = \frac{2 G M_e}{R_e} \]
Taking the square root of both sides gives us the escape velocity:
\[ V_e = \sqrt{\frac{2 G M_e}{R_e}} \]

### Final Answer
Thus, the escape velocity \( V_e \) of a sphere of mass \( m \) from the surface of the Earth is:
\[ V_e = \sqrt{\frac{2 G M_e}{R_e}} \]

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