Home
Class 12
PHYSICS
The acceleration due to gravity on the s...

The acceleration due to gravity on the surface of earth varies

Text Solution

Verified by Experts

Data : ` M_(s) = 95.22 M_(E) , R_(s) = 9.47 R_(E) .g_(E) = 9.8 m//s^(2)`
` g = (GM)/(R^(2)) `
` (g_(s))/(g_(E)) = M_(s)/M_(E) R_(E)/(R_(s))^(2)`
Surface gravity on Saturn.
` g_(s) = 9.8 xx (95.22)/((9.47)^(2))= 10.41 m//s^(2)`
Promotional Banner

Topper's Solved these Questions

  • SOLVED PROBLEMS - I

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise ROTATIONAL MOTION|26 Videos

  • SOLVED PROBLEMS - I

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise ELASTICITY|26 Videos

  • SOLVED PROBLEMS - I

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise ASSIGNMENTS|24 Videos

  • SHORT ANSWER QUESTIONS

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise Assignments|3 Videos

  • SOLVED PROBLEMS-II

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise Assignments|5 Videos

Similar Questions

Explore conceptually related problems

The ratio of acceleration due to gravity at a height 3 R above earth's surface to the acceleration due to gravity on the surface of earth is (R = radius of earth)

The ratio of acceleration due to gravity at a height 3R above earth 's surface to the acceleration due to gravity on the surface of the earth is (where R=radius of earth)

If the acceleration due to gravity on the surface of the earth is 9.8m//s^(2), what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth ?