Question

A meteor of mass 'm' having a speed 'v' at infinity reaches the surface of the earth with a speed of (\(v_e\) is escape speed from the earth's surface)

• A. \( \sqrt{2}v_e \)
• B. \( v_e \)
• C. \( 2\sqrt{v^2+v_e^2} \)
• D. \( \sqrt{v^2+v_e^2} \)

Hint:

Conservation of energy is a powerful tool for orbital mechanics and celestial body motion problems. Remember that gravitational potential energy is taken to be zero at infinity. The escape velocity formula \( v_e = \sqrt{2GM/R} \) is fundamental and frequently used.

Answer 1

The Correct Option is D

We can solve this problem using the principle of conservation of mechanical energy.
The total energy of the meteor at infinity must be equal to its total energy when it reaches the Earth's surface.
Total Energy = Kinetic Energy + Potential Energy.
At infinity (r = \(\infty\)):
The kinetic energy is \( K.E._\infty = \frac{1}{2}mv^2 \).
The gravitational potential energy is \( P.E._\infty = -\frac{GMm}{r} = 0 \), as \( r \to \infty \).
So, the total initial energy is \( E_{initial} = \frac{1}{2}mv^2 \).
At the Earth's surface (r = R, where R is the radius of the Earth):
Let the speed of the meteor be \(v_{surface}\). The kinetic energy is \( K.E._{surface} = \frac{1}{2}mv_{surface}^2 \).
The gravitational potential energy is \( P.E._{surface} = -\frac{GMm}{R} \).
So, the total final energy is \( E_{final} = \frac{1}{2}mv_{surface}^2 - \frac{GMm}{R} \).
By conservation of energy, \( E_{initial} = E_{final} \).
\( \frac{1}{2}mv^2 = \frac{1}{2}mv_{surface}^2 - \frac{GMm}{R} \).
Divide by m and multiply by 2: \( v^2 = v_{surface}^2 - \frac{2GM}{R} \).
We know that the escape speed from the Earth's surface, \(v_e\), is defined by \( \frac{1}{2}mv_e^2 = \frac{GMm}{R} \), which gives \( v_e^2 = \frac{2GM}{R} \).
Substitute this into our energy equation: \( v^2 = v_{surface}^2 - v_e^2 \).
Solving for the speed at the surface: \( v_{surface}^2 = v^2 + v_e^2 \).
\( v_{surface} = \sqrt{v^2 + v_e^2} \).