Question

If $v_e$ is escape velocity and $v _o$ is orbital velocity of a satellite for orbit close to the earth's suface, then these are related by

• A. $v_o = \sqrt 2v_e$
• B. $v_o = v_e$
• C. $v_e = \sqrt 2v_o$
• D. $v_e = \sqrt 2$

Answer 1

The Correct Option is C

The problem asks us to find the relationship between the escape velocity $v_e$ and the orbital velocity $v_o$ of a satellite for an orbit close to the Earth's surface.

To solve this, let's review the definitions of escape velocity and orbital velocity:

• Escape Velocity ($v_e$): This is the minimum velocity required to escape the gravitational influence of Earth without any further propulsion. It is given by the formula: v_e = \sqrt{2gR}
• Orbital Velocity ($v_o$): This is the velocity required to maintain a circular orbit just above the Earth's surface. It is calculated as: v_o = \sqrt{gR}

Here, $g$ is the acceleration due to gravity and $R$ is the radius of the Earth.

By substituting the formulae, we see the following relationships:

v_e = \sqrt{2gR}
v_o = \sqrt{gR}

Now, let's find the relationship between $v_e$ and $v_o$:

1. Start with $v_o = \sqrt{gR}$.
2. Multiply v_o by \sqrt{2}: \sqrt{2}v_o = \sqrt{2}\sqrt{gR} = \sqrt{2gR}.
3. Note that \sqrt{2gR} = v_e by definition.

This leads us to the equation: v_e = \sqrt{2}v_o, which matches the given correct answer.

Therefore, the escape velocity is equal to the orbital velocity multiplied by \sqrt{2}.