A projectile is fired upwards from the surface of the earth with a velocity $kv_e$ where $v_e$ is the escape velocity and $k<1$. If $r$ is the maximum distance from the center of the earth to which it rises and $R$ is the radius of the earth, then $r$ equals:
1. $\frac{R}{k^2}$
2. $\frac{R}{1-k^2}$
3. $\frac{2R}{1-k^2}$
4. $\frac{2R}{1+k^2}$

The gravitational potential difference between the surface of a planet and 10 m above is 5 J/kg. If the gravitational field is supposed to be uniform, the work done in moving a 2 kg mass from the surface of the planet to a height of 8 m is

1.  2J

2.  4J

3.  6J

4.  8J

In planetary motion, the areal velocity of the position vector of a planet depends on the angular velocity $(\omega)$ and the distance of the planet from the sun $(r)$. The correct relation for areal velocity is:
1. $\frac{dA}{dt}\propto \omega r$
2. $\frac{dA}{dt}\propto \omega^2 r$
3. $\frac{dA}{dt}\propto \omega r^2$
4. $\frac{dA}{dt}\propto \sqrt{\omega r}$

If $A$ is the areal velocity of a planet of mass $M,$ then its angular momentum is:

Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

1.  $-\frac{5\mathrm{Gm}}{\mathrm{r}}$

2.  $-\frac{6\mathrm{Gm}}{\mathrm{r}}$

3.  $-\frac{9\mathrm{Gm}}{\mathrm{r}}$

4.  0

Magnitude of potential energy ($U$) and time period $(T)$ of a satellite are related to each other as:
1. $T^2\propto \frac{1}{U^{3}}$
2. $T\propto \frac{1}{U^{3}}$
3. $T^2\propto U^3$
4. $T^2\propto \frac{1}{U^{2}}$

A projectile fired vertically upwards with a speed v escapes from the earth. If it is to be fired at 45$°$ to the horizontal, what should be its speed so that it escapes from the earth?

1.  v

2.  $\frac{\mathrm{v}}{\sqrt{2}}$

3.  $\sqrt{2}\mathrm{v}$

4.  2v

Kepler's second law regarding constancy of the areal velocity of a planet is a consequence of the law of conservation of:

1. Energy

2. Linear momentum

3. Angular momentum

4. Mass