Calculate the value of g in the following two cases:
(a) If a mango of mass `1/2` kg falls from a tree from a height of 15 metres, what is the acceleration due to gravity when it beigns to fall?
(b) Consider a satellite orbiting the Earth in a circular orbit of radius 1600 km above the surface of the Earth. What is the acceleration experienced by the satellite due to Earth's gravitational force?

Explanation

`g. = g (1-(2h)/(R_(e)))`
` g . = 9.8 (1- (2xx15)/(6400 xx10^(3))) rArr g. = 9.8 (1-0.469 xx 10^(-5))`
But `1 - 0.00000469 cong 1 `
Therefore g . = g
(b) `g. = g (1-(2h)/(R_(e)))`
` g . = g (1-(2xx1600xx10^(3))/(6400 xx10^(3))) rArr g\ = g (1-2/4)`
`g. = g (1-2/4)`
`g. = g(1-1/2)= g//2 `
The baove two examples show that the acceleration due to gravity is a constant near the surface of the Earth .