Home
Class 9
PHYSICS
One earth, value of G=6.67 xx 10^(11)Nm^...

One earth, value of `G=6.67 xx 10^(11)Nm^(2)kg^(2)`. What is its value on moon, where g is nearly `(1)/(6)`th that of earth ?

Text Solution

Verified by Experts

value of G is same on moon as on earth, because G is a universal gravitational constant.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • GRAVITATION

    PRADEEP|Exercise Value Based Questions|4 Videos

  • GRAVITATION

    PRADEEP|Exercise Problem for practice|27 Videos

  • GRAVITATION

    PRADEEP|Exercise very short answer|22 Videos

  • FORCES AND LAWS OF MOTION

    PRADEEP|Exercise MOCK TEST|36 Videos

  • MOTION

    PRADEEP|Exercise Mock test|24 Videos

Similar Questions

Explore conceptually related problems

On Earth value of G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) . What is its value on Moon, where g is nearly one-sixth than that of Earth?

Convert G=6.67xx10^(-11)Nm^(2)kg^(-2)" to "cm^(3)g^(-1)s^(-2)

The value of gravitation is G = 6.67 xx 10^(-11)N-m^(2)/kg^(2) in SL units . Convert it into CGS system of units .

The radius of the moom is 1.7 xx 10^(6) m and its mass is 7.4 xx 10^(22) kg . If G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , find the value of acceleration due to gravity on the surface of moon.

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 xx 10^(22) kg , radius = 1.75 xx 10^(6) m and gravitational constant = 6.67 xx 10^(-11) N m^(2) kg^(-2) . What will be the mass of the person at the Moon and acceleration due to gravity there ? If this person can jump 2 m high on the Earth, how much high can he jump at the Moon ?

A satellite orbits the earth at a height of 500 km from its surface. Compute its (i) Kinetic energy, (ii) potential energy, and (iii) total energy. Mass of the satellite=300 kg, Mass of the earth= 6.0xx10^(24) kg, radius of the earth= 6.4xx10^(6) m, G= 6.67xx10^(-11)Nm^(2)Kg^(-2) . Will your answer alter if the earth were to shrink suddenly to half its size?

Convert G=6.67xx10^(-11)Nm^(2)kg^(-2) to cm^(3)s^(-2)g^(-1) , where G is the universal gravitational constant.

If a spaceship orbits the earth at a height of 500 km from its surface, then determine its (i) kinetic energy, (ii) potential energy, and (iii) total energy (iv) binding energy. Mass of the satellite = 300 kg, Mass of the earth = 6xx10^(24) kg , radius of the earth = 6.4 xx 10^(6) m, G=6.67 xx 10^(-11) "N-m"^(2) kg^(-2) . Will your answer alter if the earth were to shrink suddenly to half its size ?

Calculate the mass of the sun if the mean orbital radius of Jupiter around the sun is 7.8 xx 10^(11) m . Take G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) Time period of revolution = 12 years according to earth

What is the value of gravitational constant G=6.67xx10^(-11)Nm^(2)//kg^(2) in CGS units?

Home
Profile