Home
Class 11
PHYSICS
If 'g' on the surface of the earth is 9....

If 'g' on the surface of the earth is `9.8ms^(-2)`, find its value at a depth of 3200 km (radius of the earth = 6400 km)

Text Solution

AI Generated Solution

To find the value of acceleration due to gravity (g) at a depth of 3200 km below the surface of the Earth, we can use the formula: \[ g_d = g \left(1 - \frac{d}{r}\right) \] where: - \( g_d \) is the acceleration due to gravity at depth, - \( g \) is the acceleration due to gravity on the surface (given as 9.8 m/s²), - \( d \) is the depth (3200 km), ...
Promotional Banner

Similar Questions

Explore conceptually related problems

If g on the surface of the earth is 9.8 m//s^(2) , find its value at a height of 6400 km. (Radius of the earth = 6400 km)

If the value of g at the surface of the earth is 9.8 m//sec^(2) , then the value of g at a place 480 km above the surface of the earth will be (Radius of the earth is 6400 km)

Find out the capacitance of the earth ? (Radius of the earth = 6400 km)

A satellite is a at height of 25, 600 km from the surface of the earth. If its orbital speed is 3.536 km/s find its time period. (Radius of the earth = 6400 km)

Value of g on the surface of earth is 9.8 m//s^(2) . Find its value on the surface of a planet whose mass and radius both are two times that of earth.

A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is 6400 km.)

What is the angular velocity of a body on'the surface of the earth at the equator ? Also find its linear velocity. Given radius of the earth is 6400 km. Period of rotation of the earth = 24 hours.

Find the ratio of the orbital speeds of two satellites of the earth if the satellites are at height 6400 km and 19200 km. (Radius of the earth = 6400 km).

Value of g on the surface of earth is 9.8 m//s^(2) . Find its (a) at height h = R from the surface , (b) at depth d = (R)/(2) from the surface . ( R = radius of earth)

A what height above the earth's surface the value of g becomes 25% of its value on the earth if radius of the earth is 6400km.