the acceleration due to gravity becomes g/2 where g =acceleration due to gravity on the surface of the earth at a height equal to
Answers
Answered by
0
F = G ( mM/R²). This is the magnitude of the force. Its direction is directed along a radius of the earth from the centre of the smaller body to the centre of the earth. This force produces an acceleration g at/near the earth's surface such that;
mg = G ( m M/ R²); ==> g = G M/R² ———————(1).
The acceleration due to gravity g’ at a height h from the surface of the earth is,
g' = G M/(R + h)² = G M /[ R (1 +(h/R)]²;
==> g' = ( GM/R²)[ 1 + h/R]^-2 = g ( 1 - h/R)^-2
==> g'/g = ( 1 - h/R)^-2 = ( 1 - 2 h/ R; ——-—————-(2)
by solving (1) & (2) u will get your answer
mg = G ( m M/ R²); ==> g = G M/R² ———————(1).
The acceleration due to gravity g’ at a height h from the surface of the earth is,
g' = G M/(R + h)² = G M /[ R (1 +(h/R)]²;
==> g' = ( GM/R²)[ 1 + h/R]^-2 = g ( 1 - h/R)^-2
==> g'/g = ( 1 - h/R)^-2 = ( 1 - 2 h/ R; ——-—————-(2)
by solving (1) & (2) u will get your answer
Similar questions