Gravitation · Physics · COMEDK
MCQ (Single Correct Answer)
The acceleration due to gravity at pole and equator can be related as
A satellite is revolving around the earth in a circular orbit with kinetic energy of $$1.69 \times 10^{10} \mathrm{~J}$$. The additional kinetic energy required for just escaping into the outer space is
A planet has double the mass of the earth and double the radius. The gravitational potential at the surface of the Earth is $$\mathrm{V}$$ and the magnitude of the gravitational field strength is $$\mathrm{g}$$. The gravitational potential and gravitational field strength on the surface of the planet are
| Potential | Field | |
|---|---|---|
| A | V | $$\frac{g}{4}$$ |
| B | 2V | $$\frac{g}{2}$$ |
| C | V | $$\frac{g}{2}$$ |
| D | 2V | $$\frac{g}{4}$$ |
Energy required for moving a body of mass $$\mathrm{m}$$ from a circular orbit of radius 3R to a higher orbit of radius 4R around the earth is.
If $$\mathrm{A}$$ is the areal velocity of a planet of mass $$\mathrm{M}$$, then its angular momentum is
If the earth has a mass nine times and radius four times that of planet X, the ratio of the maximum speed required by a rocket to pull out of the gravitational force of planet $$\mathrm{X}$$ to that of the earth is
Starting from the centre of the earth having radius $$R$$, the variation of $$g$$ (acceleration due to gravity) is shown by
If escape velocity on earth surface is $$11.1 \mathrm{~kmh}^{-1}$$, then find the escape velocity on moon surface. If mass of moon is $$\frac{1}{81}$$ times of mass of earth and radius of moon is $$\frac{1}{4}$$ times radius of earth.
The height vertically above the earth's surface at which the acceleration due to gravity becomes $$1 \%$$ of its value at the surface is
An uniform sphere of mass $$M$$ and radius $$R$$ exerts a force of $$F$$ on a small mass $$m$$ placed at a distance of 3R from the centre of the sphere. A spherical portion of diameter $$R$$ is cut from the sphere as shown in the fig. The force of attraction between the remaining part of the disc and the mass $$\mathrm{m}$$ is
The acceleration due to gravity at a height of $$7 \mathrm{~km}$$ above the earth is the same as at a depth d below the surface of the earth. Then d is
If the earth were to spin faster, acceleration due to gravity at the poles
The height at which the acceleration due to gravity becomes $$\frac{g}{16}$$ (where, g = acceleration due to gravity on the surface of the earth) in terms of R is, if R is the radius of earth.
The escape velocity of a projectile on the earth's surface is 11.2 km/s. A body is projected out with thrice this speed. The speed of the body far away from the earth will be
Kepler's second law of planetary motion corresponds to
A constant potential energy of a satellite is given as
$$\mathrm{PE}=r(\mathrm{KE})$$
whee, PE = potential energy
and KE = kinetic energy.
The value of $$r$$ will be
A satellite can be in a geostationary orbit around a planet if it is at a distance R from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital radius should be
Two spherical bodies of masses M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is