## Question

The intensity of gravitational field at a point situated at earth’s surface is 2.5 N/kg. Calculate the gravitational potential at that point. Given: radius of earth, .

### Solution

–1.6 × 10^{7} J/kg

The whole mass of the earth is assumed to be concentrated at its centre. Hence the intensity of gravitational field at the earth’s surface is given by

Where *M _{e}* and

*R*are respectively mass and radius of the earth. The gravitational potential at the earth’s surface is

_{e}V = - GM_{e}/R_{e}

Substituting the given values:

N-m/kg = –1.6 × 10^{7} J/kg

#### SIMILAR QUESTIONS

Two masses, 800 kg and 600 kg, are at a distance 0.25 m apart. Compute the magnitude of the intensity of the gravitational field at a point distant 0.20 m from the 800 kg mass and 0.15 m from the 600 kg mass (*G* = 6.66× 10^{ –11} N m^{2} kg^{–2}).

Three particles, each of mass *m*, are situates at the vertices of an equilateral triangle of side length *a*. The only force acting on the particles are their mutual gravitational forces. It is desired that each particle move in a circle while maintaining the original mutual separation *a*. Find the initial velocity that should be given to each particle and also the time-period of the circular motion.

The weight of a person on the earth is 80 kg. What will be his weight on the moon? Mass of the moon = 7.34 × 10^{22}kg, radius = 1.75 × 10^{6} m and gravitational constant *G* = 6.67 × 10^{ –11} Nm^{2}/kg^{2}. What will be the mass of the person at the moon? If this person can jump 2 meter high on the earth, how much high can he jump at the moon? If he can walk 100 m in 1 minute on the earth, then how much will he walk in 1 minute on the moon?

What will be the acceleration due to gravity on the surface of the moon if its radius is 1/4^{th} the radius of the earth and its mass is (1/80)th the mass of the earth.

Calculate that imaginary angular velocity of the earth for which effective acceleration due to gravity at the equator becomes zero. In this condition what will be the length (in hours) of the day?

(*R _{e}* = 6400 km,

*g*= 10 ms

^{–2})

Determine the speed with which the earth would have to rotate on its axis so that a person on the equator would weigh 3/5 th as much as at present. Take the equatorial radius as 6400 km.

At what height above the earth’s surface the acceleration due to gravity will be 1/9 th of its value at the earth’s surface? Radius of earth is 6400 km.

The radius of the earth is approximately 6000 km. What will be your weight at 6000 km above the surface of the earth? At 12000 km above? At 18000 km above?

Calculate the gravitational field strength and the gravitational potential at the surface of the moon. The mass of the moon is kg and its radius is

.

At a point above the surface of the earth, the gravitational potential is and the acceleration due to gravity is 6.4 ms^{–2}. Assuming the mean radius of the earth to be 6400 km, calculate the height of this point above the earth’s surface. ^{ }