For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of:

1.	\(2\)	2.	\(\sqrt{2}\)
3.	\(1/\sqrt{2}\)	4.	\(4\)

The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes \(1\%\) of its value at the surface is: (Radius of the earth = \(R\))
1. \(8R\)
2. \(9R\)
3. \(10R\)
4. \(20R\)

If the density of the earth is increased \(4\) times and its radius becomes half of what it is, our weight will be:
1. four times the present value
2. doubled
3. the same
4. Halved

The escape velocity for the Earth is taken \(v_d\). Then, the escape velocity for a planet whose radius is four times and the density is nine times that of the earth, is:

The value of \(g\) at a particular point is \(9.8~\text{m/s}^2\). Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass then value of \(g\) at the same point will now become: (assuming that the distance of the point from the centre of the earth does not shrink)

The change in the potential energy, when a body of mass \(m\) is raised to a height \(nR\) from the Earth's surface is: (\(R\) = Radius of the Earth)
1. \(mgR\left(\frac{n}{n-1}\right)\)
2. \(nmgR\)
3. \(mgR\left(\frac{n^2}{n^2+1}\right)\)
4. \(mgR\left(\frac{n}{n+1}\right)\)

A satellite whose mass is \(m\), is revolving in a circular orbit of radius \(r\), around the earth of mass \(M\). Time of revolution of the satellite is:
1. \(T \propto \frac{r^5}{GM}\)
2. \(T \propto \sqrt{\frac{r^3}{GM}}\)
3. \(T \propto \sqrt{\frac{r}{\frac{GM^2}{3}}}\)
4. \(T \propto \sqrt{\frac{r^3}{\frac{GM^2}{4}}}\)

Suppose the gravitational force varies inversely as the ${\mathrm{}}^{}$\(n^{th}\) power of distance then the time period of a planet in circular orbit of radius \(R\) around the sun will be proportional to:
1. \(R^{\left(\frac{n+1}{2}\right)}\)
2. \(R^{\left(\frac{n-1}{2}\right)}\)
3. \(R^n\)
4. \(R^{\left(\frac{n-2}{2}\right)}\)

Time period of a satellite revolving above Earth’s surface at a height equal to \(R\) (the radius of Earth) will be:
(\(g\) is the acceleration due to gravity at Earth’s surface)
1. \(2 \pi \sqrt{\frac{2 R}{g}}\)
2. \(4 \sqrt{2} \pi \sqrt{\frac{R}{g}}\)
3. \(2 \pi \sqrt{\frac{R}{g}}\)
4. \(8 \pi \sqrt{\frac{R}{g}}\)