Question

An asteroid of mass $m$ is approaching earth, initially at a distance $10{R_E}$ with speed ${v_i}$ . it hits earth with a speed ${v_f}$ ( ${R_E}$ and ${M_E}$ are the radius and mass of the earth), then
(A) ${v_f}^2 = {v_i}^2 + \dfrac{{2Gm}}{{{R_E}}}\left( {1 + \dfrac{1}{{10}}} \right)$
(B) ${v_f}^2 = {v_i}^2 + \dfrac{{2G{M_E}}}{{{R_E}}}\left( {1 + \dfrac{1}{{10}}} \right)$
(C) ${v_f}^2 = {v_i}^2 + \dfrac{{2G{M_E}}}{{{R_E}}}\left( {1 - \dfrac{1}{{10}}} \right)$
(D) ${v_f}^2 = {v_i}^2 + \dfrac{{2Gm}}{{{R_E}}}\left( {1 - \dfrac{1}{{10}}} \right)$

Answer 1

Hint : Use the equation for kinetic energy and potential energy for a body in the gravitational field and law of conservation of energy to find the final velocity of the asteroid . The kinetic energy of a body is given by, $K.E = \dfrac{1}{2}m{v^2}$ and potential energy of a body (or the system) in gravitational field with distance between the centre of the bodies $R$ is $U = - \dfrac{{GMm}}{R}$ where, $M$ is the mass of the other body giving the mass $m$ a gravitational force $\dfrac{{GMm}}{{{R^2}}}$ , and $G$ is the gravitational constant.

Complete Step By Step Answer:
We know that the total energy of a body in a conservative force field is constant. Since, we know the gravitational force field is a conservative field, hence, the total energy of the asteroid is constant. Hence, energy of the asteroid at $10{R_E}$ is equal to the energy of the asteroid at ${R_E}$ (when hitting the earth)
Now, the initial energy of the asteroid is, ${E_i} = {K_i} + {U_i} = \dfrac{1}{2}mv_i^2 - \dfrac{{G{M_E}m}}{{10{R_E}}}$ where, $m$ is the mass of the asteroid, ${R_E}$ and ${M_E}$ are the radius and mass of the earth respectively. ${v_i}$ is the velocity of the asteroid at $10{R_E}$ .
The final energy of the asteroid when it hits earth is,
${E_f} = \dfrac{1}{2}mv_f^2 - \dfrac{{G{M_E}m}}{{{R_E}}}$ where, ${v_f}$ is the final velocity of the asteroid just before hitting the earth at ${R_E}$ .
Since, the total energy is constant, hence,
${E_i} = {E_f}$
Therefore, $\dfrac{1}{2}mv_i^2 - \dfrac{{G{M_E}m}}{{10{R_E}}} = \dfrac{1}{2}mv_f^2 - \dfrac{{G{M_E}m}}{{{R_E}}}$
Or, $v_f^2 - \dfrac{{2G{M_E}}}{{{R_E}}} = v_i^2 - \dfrac{{2G{M_E}}}{{10{R_E}}}$
Or, $v_f^2 = v_i^2 + \dfrac{{2G{M_E}}}{{{R_E}}}\left( {1 - \dfrac{1}{{10}}} \right)$
Hence, the relation between the final and initial velocity of the asteroid is given by, $v_f^2 = v_i^2 + \dfrac{{2G{M_E}}}{{{R_E}}}\left( {1 - \dfrac{1}{{10}}} \right)$
Hence, option ( C) is correct.

Note :
From the relation one can observe that the initial velocity depends on the radius and mass of the earth; only the initial velocity of the asteroid is constant.
If the acceleration of the earth due to the asteroid is large enough (comparable to the sun's force of attraction ) then the earth can move towards the asteroid and collide with each other also.