# Iif the acceleration above the earth is is gMm/[r+h] then acceleration due to gravity below earth is?

1
by dippy9

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by dippy9

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Your formula is wrong.

Gravity above Earth's surface is g' = G Me / (r+h)²

Force of gravity = m g'

See the diagram enclosed.

When the mass m is below earth's surface at a depth of h , then there are two forces acting on it. One is the gravitational force of Earth inside radius (R-h) and another, the gravity due to Shell of thickness h.

The net force of attraction of the outer shell of thickness h is zero. This is Because of symmetry and direction of force being spread in 360 degrees from all sides.

Density of Earth, = Me / (4πR³/3)

Mass of Earth with in radius (R-h) = Me * [ 4π(R-h)³/3 ] / (4πR³/3)

Me' = Me (R-h)³ / R³

Hence the gravity (acceleration) g' below Earth's surface =

g' = G Me' / (R-h)² = G Me (R-h) / R³

= ( GMe / R² ) [ 1 - h / R ]

g' = g [ 1 - h/R ]

m g' = m g ( 1 - h/R )

Gravity above Earth's surface is g' = G Me / (r+h)²

Force of gravity = m g'

See the diagram enclosed.

When the mass m is below earth's surface at a depth of h , then there are two forces acting on it. One is the gravitational force of Earth inside radius (R-h) and another, the gravity due to Shell of thickness h.

The net force of attraction of the outer shell of thickness h is zero. This is Because of symmetry and direction of force being spread in 360 degrees from all sides.

Density of Earth, = Me / (4πR³/3)

Mass of Earth with in radius (R-h) = Me * [ 4π(R-h)³/3 ] / (4πR³/3)

Me' = Me (R-h)³ / R³

Hence the gravity (acceleration) g' below Earth's surface =

g' = G Me' / (R-h)² = G Me (R-h) / R³

= ( GMe / R² ) [ 1 - h / R ]

g' = g [ 1 - h/R ]

m g' = m g ( 1 - h/R )