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

A point \(P\) lies on the axis of a ring of mass \(M\) and radius \(a\) at a distance \(a\) from its centre \(C\). A small particle starts from \(P\) and reaches \(C\) under gravitational attraction. Its speed at \(C\) will be:
1. \(\sqrt{\frac{2 GM}{a}}\)
2. \(\sqrt{\frac{2 GM}{a} \left(1 - \frac{1}{\sqrt{2}}\right)}\)
3. \(\sqrt{\frac{2 GM}{a} \left(\sqrt{2} - 1\right)}\)
4. zero

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

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. \(T^2\propto \frac{1}{U^{2}}\)

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

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

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. \(\frac{dA}{dt}\propto \sqrt{\omega r}\)

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

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. \(\frac{2R}{1+k^2}\)