Find the value of acceleration due to gravity at the height of 80 km from the surface of earth of radius 6400 km when the value of acceleration due to gravity on the surface of earth is 10 M per second
Answers
Newtons law of Gravitation states that the force of gravitational attraction between two bodies of mass M and m separated by a distance d, is proportional directly to the product of masses and inversely proportional to the square of the distance between them.
Newton's law of Gravitation: F = G M m / d²
G = Universal Gravitational constant = 6.674 * 10⁻¹¹ N m²/kg²
M = Mass of a body
m = mass of the 2nd body
d = distance between the centers of mass of the two bodies
Let R = radius of Earth = 6,400 km
h = altitude above Earth = 12,800 km
d = distance from center of Earth = 19,200 km
We know that on the surface of Earth, gravity is:
g = G M / R² = 9.81 m/s²
g' = acceleration due to gravity at distance d = G M / d²
=> g'/g = R²/d²
=> g' = g R² / d²
= 9.81 * 6400²/19200²
= 1.09 m/s²
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