91Ó°ÊÓ

1. The figure for three uniform identical spherical planets is given.
2. The periods of rotation for three planets are given in the figure.

The acceleration of free fall is the acceleration at which a body falling freely experiences the gravitational pull of the earth alone. We can find the relation between the value of free-fall acceleration g and the period of rotation of the planet, T. By comparing these periods, we can rank the points on the planet according to the corresponding value of free-fall acceleration g.

Formula:

The gravitational acceleration of the body, $a_g=\frac{GM}{R^2}$ …(¾±)

The effect of the rotation of Earth on the acceleration due to gravity, $g=a_g-{\mathrm{Ó\neg }}^2\mathrm{R}$ …(¾±¾±)

The angular velocity of a body in rotation, $\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$ …(¾±¾±¾±)

The effect of rotation of Earth on the acceleration due to gravity can be given using equations (i) and (iii) in equation (ii) as follows:

$g=\frac{G{M}^2}{R^2}-\frac{4{\mathrm{Ï€}}^2}{T^2}R$

Since the planets are identical, the value of free-fall acceleration g will increase with increasing period of rotation of the planet. This is because as the time period increases, the second term in the above equation would become smaller and smaller. Hence g would increase.

The points b, d and f which are on the pole which do not orbit. Their period of rotation is zero.

So, the value of free-fall acceleration g at point b, d and f is maximum.

The period of rotation of the planet at point e is greatest. So, the value of free-fall acceleration g at point e is greater than c and a but smaller than b, d and f.

The period of rotation of the planet at point a is the least. So, the value of free-fall acceleration g at point a is the smallest.

Therefore, the ranking of the points according to the corresponding value of free-fall acceleration g at them, greatest first is $b=d=f>e>c>a$.