If the radius of a planet is \(\mathrm{R}\) and its density is , the escape velocity from its surface will be:
1.
2.
3.
4.
The escape velocity for a rocket from the earth is \(11.2\) km/s. Its value on a planet where the acceleration due to gravity is double that on the earth and the diameter of the planet is twice that of the earth (in km/s) will be:
1. | \(11.2\) | 2. | \(5.6\) |
3. | \(22.4\) | 4. | \(53.6\) |
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:
1. | \(36v_d\) | 2. | \(12v_d\) |
3. | \(6v_d\) | 4. | \(20v_d\) |
For a planet having mass equal to the mass of the earth but radius equal to one-fourth of the radius of the earth, its escape velocity will be:
1. | 11.2 km/s | 2. | 22.4 km/s |
3. | 5.6 km/s | 4. | 44.8 km/s |
The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fve, where ve is its escape velocity from the surface of the earth. The value of f is:
1.
2.
3.
4.
A projectile is fired upwards from the surface of the earth with a velocity where 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. \(\frac{2R}{1+k^2}\)
A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is:
1. R/3
2. R/2
3. R/4
4. R/5
The initial velocity \(v_i\) required to project a body vertically upwards from the surface of the earth to just reach a height of \(10R\), where \(R\) is the radius of the earth, described in terms of escape velocity \(v_e\) is:
1. \(\sqrt{\frac{10}{11}}v_e\)
2. \(\sqrt{\frac{11}{10}}v_e\)
3. \(\sqrt{\frac{20}{11}}v_e\)
4. \(\sqrt{\frac{11}{20}}v_e\)
A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would Earth (mass \(= 5.98\times 10^{24}~\text{kg}\)) have to be compressed to be a black hole?
1. \(10^{-9}~\text{m}\)
2. \(10^{-6}~\text{m}\)
3. \(10^{-2}~\text{m}\)
4. \(100~\text{m}\)
A satellite is revolving in a circular orbit at a height \(h\) from the earth's surface (radius of earth \(R\); \(h<<R\)). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field is close to: (Neglect the effect of the atmosphere.)
1. \(\sqrt{2gR}\)
2. \(\sqrt{gR}\)
3. \(\sqrt{\frac{gR}{2}}\)
4. \(\sqrt{gR}\left(\sqrt{2}-1\right)\)