From what height above the surface of the earth should an object be dropped to initially experience an acceleration of 0.54g? (G = 6.67 × 10-11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg, Rearth = 6.38 × 106 m)

Respuesta :

In order to solve the problem, it is necessary to apply the concepts related to Acceleration due to gravity.

By general definition we know that gravity is defined under the Newtonian formula:

[tex]g=\frac{GM}{R^2}[/tex]

Where,

G= Gravitational Constant

M = Mass

R = Radius

When a height of a body is increased it is necessary to increase that height in the given Radius, therefore,

[tex]\frac{GM}{(R+h)^2}=0.54g[/tex]

Replacing,

[tex]\frac{(6.67*10^{-11})(5.97*10^{24})}{(6.38*10^6+h)^2}=0.54g[/tex]

[tex]\frac{(6.67*10^{-11})(5.97*10^{24})}{0.54g}=(6.38*10^6+h)^2[/tex]

[tex]\sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{0.54g}}=(6.38*10^6+h)[/tex]

[tex]h= \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{0.54g}}-6.38*10^6[/tex]

[tex]h=2.29441*10^6m[/tex]

Therefore the height above the surface of the earth should be 2300Km