The planet Mars has two moons, Phobos and Delmos. Phobos has a period of \(7\) hours, \(39\) minutes and an orbital radius of $9.4\times {10}^3$ km. The mass of mars is:
1. $6.48\times {10}^{23} \mathrm{kg}$
2. $6.48\times {10}^{25} \mathrm{kg}$
3. $6.48\times {10}^{20} \mathrm{kg}$
4. $6.48\times {10}^{21} \mathrm{kg}$

Assume that a space shuttle flies in a circular orbit very close to the earth's surface. Taking the radius of the space shuttle's orbit to be equal to the radius of the earth (\(R\)) and the acceleration due to gravity to be \(g\), the time period of one revolution of the space shuttle is (nearly):

A remote sensing satellite of earth revolves in a circular orbit at a height of \(0.25 \times10^6~\text{m}\) above the surface of the earth. If Earth’s radius is \(6.38\times10^6~\text{m}\) and \(g=9.8~\text{ms}^{-2}\), then the orbital speed of the satellite is:
1. \(7.76~\text{kms}^{-1}\)
2. \(8.56~\text{kms}^{-1}\)
3. \(9.13~\text{kms}^{-1}\)
4. \(6.67~\text{kms}^{-1}\)

The radii of circular orbits of two satellites A and B of the earth are \(4R\) and \(R\) respectively. If the speed of satellite A is \(3v,\) then the speed of satellite B will be:
1. \(3v/4\)
2. \(6v\)
3. \(12v\)
4. \(3v/2\)

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is:
1. $\frac{1}{\sqrt{2}}$
2. \(2\)
3. $\sqrt{2}$
4. $\frac{1}{2}$

The time period of an earth satellite in circular orbit is independent of:

Two satellites A and B move around the earth in the same orbit. The mass of B is twice the mass of A. Then: