Suppose,the acceleration due to gravity at the earth's surface is $10 \, m/s^2$ and at the surface of Mars it is $4.0 \, m/s^2$. $A$ $60 \, kg$ passenger goes from the earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of the figure best represents the weight (net gravitational force) of the passenger as a function of time?

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 $\frac{\Delta F_1}{\Delta F_2}$ is

The distance between the Sun and the Earth is $1.6 \times 10^{11} \,m$ and the radius of the Earth is $6.4 \times 10^6 \,m$. The ratio of the angular momentum of the Earth around the Sun to the angular momentum around its own axis is approximately (Assume the Earth as a solid sphere with uniform mass density and that it rotates around the Sun in a circular path.)

Two identical thin uniform rods of mass $M$ and length $L$ are placed in a line at a separation of $L$. Find the gravitational force acting between them.

$Assertion$ : The Earth is slowing down and as a result the Moon is coming nearer to it.
$Reason$ : The angular momentum of the Earth-Moon system is not conserved.

As shown in the figure,two spherical cavities are made in a uniform solid sphere of radius $R$. The boundaries of the cavities touch at the centre of the sphere. The centers of the cavities and the sphere lie on the $X$-axis. The mass of the solid sphere before the cavities were created was $M$. The gravitational force on a point mass $m$ at a distance $d$ away from the centre of the solid sphere is