Question
A spherical planet has a mass $${M_p}$$ and diameter $${D_p}.$$ A particle of mass $$m$$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to
A.
$$\frac{{4G{M_p}}}{{D_p^2}}$$
B.
$$\frac{{G{M_p}m}}{{D_p^2}}$$
C.
$$\frac{{G{M_p}}}{{D_p^2}}$$
D.
$$\frac{{4G{M_p}m}}{{D_p^2}}$$
Answer :
$$\frac{{4G{M_p}}}{{D_p^2}}$$
Solution :
Apply Newton's gravitation law. According to Newton's law of gravitation force, $$F = \frac{{GMm}}{{{R^2}}}$$
Force on planet of mass $${M_p}$$ and body of mass $$m$$ is given by
$$\eqalign{
& F = \frac{{G{M_p}m}}{{{{\left( {\frac{{{D_p}}}{2}} \right)}^2}}}\,\,\left[ {{\text{where,}}\,{D_p} = {\text{diameter of planet and }}{R_p} = {\text{radius of planet}} = \frac{{{D_p}}}{2}} \right] \cr
& F = \frac{{4G{M_p}m}}{{D_p^2}} \cr} $$
As we know that, $$F = ma$$
So, acceleration due to gravity $$a = \frac{F}{m} = \frac{{4G{M_p}}}{{D_p^2}}$$