The escape velocity from the Earth's surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and the same mass density is: 

1.	\(3v\)	2.	\(4v\)
3.	\(v\)	4.	\(2v\)

A particle of mass \(m\) is projected with a velocity, \(v=kV_{e} ~(k<1)\) from the surface of the earth. The maximum height, above the surface, reached by the particle is: (Where \(V_e=\) escape velocity, \(R=\) radius of the earth)

Assume that earth and mars move in circular orbits around the sun, with the martian orbit being \(1.52\) times the orbital radius of the earth. The length of the martian year in days is approximately:
(Take \((1.52)^{3/2}=1.87\))
1. \(344\) days
2. \(684\) days
3. \(584\) days
4. \(484\) days

A 400 kg satellite is in a circular orbit of radius $2{\mathrm{R}}_{\mathrm{E}}$ about the Earth. What are the changes in the kinetic and potential energies respectively to transfer it to a circular orbit of radius $4{\mathrm{R}}_{\mathrm{E}}.$ (where ${\mathrm{R}}_{\mathrm{E}}$ is the radius of the earth)

1. $3.13\times {10}^9 \mathrm{J} \mathrm{and} 6.25\times {10}^9 \mathrm{J}$

2. $3.13\times {10}^9 \mathrm{J} \mathrm{and} -6.25\times {10}^9 \mathrm{J}$

3. $-3.13\times {10}^9 \mathrm{J} \mathrm{and} -6.25\times {10}^9 \mathrm{J}$

4. $-3.13\times {10}^8 \mathrm{J} \mathrm{and} -6.25\times {10}^8 \mathrm{J}$

A \(400\) kg satellite is in a circular orbit of radius \(2R_E\) (where \(R_E\) is the radius of the earth) about the Earth. How much energy is required to transfer it to a circular orbit of radius \(4R_E\)\(?\) (Given \(R_E=6.4\times10^{6}\) m)
${\mathrm{}}_{}$1. \(3.13\times10^{9}\) J
2. \(3.13\times10^{10}\) J
3. \(4.13\times10^{9}\) J
4. \(4.13\times10^{8}\) J

Constant $\mathrm{k} = {10}^{-13} {\mathrm{s}}^2{\mathrm{m}}^{-3}$ in days and kilometres is?

1.   ${10}^{-13} {\mathrm{d}}^2{\mathrm{km}}^{-3}$

2.   $1.33\times {10}^{14} {\mathrm{dkm}}^{-3}$

3.   ${10}^{-13} {\mathrm{d}}^2\mathrm{km}$

4.   $1.33\times {10}^{-14} {\mathrm{d}}^2{\mathrm{km}}^{-3}$

You are given the following data: g = 9.81 $\mathrm{m}/{\mathrm{s}}^2$, ${\mathrm{R}}_{\mathrm{E}} = 6.37\times {10}^6$ m, the distance to the moon, R = $3.84\times {10}^8$ m and the time period of the moon’s revolution is 27.3 days. Mass of the Earth ${\mathrm{M}}_{\mathrm{E}}$ in two different ways is:

1.   $5.97\times {10}^{24} \mathrm{kg} \mathrm{and} 6.02\times {10}^{24} \mathrm{kg}$

2.   $5.97\times {10}^{24} \mathrm{kg} \mathrm{and} 6.02\times {10}^{23} \mathrm{kg}$

3.   $5.97\times {10}^{23} \mathrm{kg} \mathrm{and} 6.02\times {10}^{24} \mathrm{kg}$

4.   $5.97\times {10}^{23} \mathrm{kg} \mathrm{and} 6.02\times {10}^{23} \mathrm{kg}$

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. $6.48\times {10}^{21} \mathrm{kg}$