A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the objects :

Engineering

Physics

Gravitational Force and Field

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Circular Motion in Vertical Plane

Question

A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the objects :

$\frac{3}{2} {mV}^{2}$

2mV²

$\frac{1}{2} {mV}^{2}$

mV²

$\Rightarrow \frac{{mv}^{2}}{r} = \frac{{GmM}_{e}}{r^{2}} \Rightarrow r = \frac{{GM}_{e}}{V^{2}}$ ........(1)

If K.E. of mass m = was k then from

$E = K - \frac{{GmM}_{e}}{r} = 0 \Rightarrow K = m (\frac{{GM}_{e}}{r}) = {mv}^{2}$

This question involves understanding orbital mechanics and escape velocity. Let's break it down step by step:

Step 1: Understand the satellite's motion
The satellite is in a circular orbit around Earth with constant speed V. For a circular orbit, the centripetal force is provided by gravity:

\frac{G M m}{r^{2}} = \frac{m V^{2}}{r}

where G is the gravitational constant, M is Earth's mass, m is mass of object, and r is orbital radius.

Step 2: Simplify the orbital equation
Canceling m and r from both sides:

\frac{G M}{r} = V^{2}

This gives us the relationship between orbital speed and orbital radius.

Step 3: Recall the escape velocity formula
The escape velocity from a distance r from Earth's center is:

v_{esc} = \sqrt{\frac{2 G M}{r}}

Step 4: Relate escape velocity to orbital velocity
From Step 2, we know that $V^{2} = \frac{G M}{r}$ , so:

v_{esc} = \sqrt{2 V^{2}} = \sqrt{2} V

Step 5: Calculate the required kinetic energy
For the object to just escape, it needs kinetic energy equal to:

K E = \frac{1}{2} m v_{esc}^{2} = \frac{1}{2} m {(\sqrt{2} V)}^{2} = \frac{1}{2} m (2 V^{2}) = m V^{2}

Final Answer: The kinetic energy of the object at the time of ejection should be $m V^{2}$ .

Key Formulae

Orbital velocity: $V = \sqrt{\frac{G M}{r}}$

Escape velocity: $v_{esc} = \sqrt{\frac{2 G M}{r}}$

Kinetic energy: $K E = \frac{1}{2} m v^{2}$

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