Answer:
[tex]h = 2.36 X 10^{8}[/tex] m
Explanation:
An object kept in height h in space above the earth's surface is acted upon by force F which is given by;
F = [tex]\frac{GMm}{(R+h)^{2} }[/tex]
where F = mg
mg = [tex]\frac{GMm}{(R+h)^{2} }[/tex]
g = [tex]\frac{GM}{(R+h)^{2} }[/tex]
[tex](R+h)^{2} = \frac{GM}{g}[/tex]
M = Mass of earth = [tex]5.98 X 10^{24}[/tex] kg
R = Radius of earth = [tex]6.38 X 10^{6}[/tex] m
G = Gravitational constant = [tex]6.67259 X 10^{-11}[/tex] N[tex]m^{2}[/tex] /[tex]kg^{2}[/tex]
g = 1/140 g = [tex]7.14286 X 10^{-3}[/tex]
[tex]h^{2} = \frac{GM}{g} - R^{2}[/tex]
substituting for M, R, G and g
(Kindly ignore Armstrong symbol in the following equation, I don't know how it popped up using the equation tool)
[tex]h^{2} = \frac{(6.67259X10^{-11}) X (5.98 X 10^{24} ) }{(7.14286 X 10^{-3}) } - (6.38 X 10^{6}) ^{2}[/tex]
[tex]h^{2} = \frac{3.99 X 10^{14} }{7.14286 X 10^{-3} } - 4.07 X 10^{13}[/tex]
[tex]h^{2} = 5.586 X 10^{16} - 4.07 X 10^{13}[/tex]
[tex]h^{2} = 5.58193 X 10^{16}[/tex]
[tex]h = \sqrt{(5.58193 X 10^{16})}[/tex]
[tex]h = 2.36 X 10^{8}[/tex] m
