Take the mean distance of the moon and the sun from the earth to be $$0.4 \times {10^6}$$ km and $$150 \times {10^6}$$ km respectively. Their masses are $$8 \times {10^{22}}$$ kg and $$2 \times {10^{30}}$$ kg respectively. The radius of the earth is $$6400$$ km. Let $$\Delta {F_1}$$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $$\Delta {F_2}$$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $${{\Delta {F_1}} \over {\Delta {F_2}}}$$ is :

The mass density of a spherical body is given by
$$\rho $$ (r) = $${k \over r}$$ for r $$ \le $$ R and $$\rho $$ (r) = 0 for r > R,

where r is the distance from the centre.

The correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is :

If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $${3 \over 4}$$ W. Radius of the Earth is 6400 km and g=10 m/s².

The variation of acceleration due to gravity $$g$$ with distance d from centre of the earth is best represented by (R = Earth’s radius):