Respuesta :
Answer:
The answer to the question is
At an altitude of 1413 km or 1.222·R above the north pole the weight of an object reduced to 67% of its earth-surface value
Explanation:
We make use of the gravitational formula as follows
F = G[tex]\frac{m_{1} m_{2} }{R^{2} }[/tex] where
m₁ = mass of the object
m₂ = mass of the earth
d = distance between the two objects and
G = gravitational constant
if at the altitude the weight is reduced to 67 % of its weight on earth then
with all other variables remaining constant, we have
67% F = G[tex]\frac{m_{1} m_{2} }{R_{2} ^{2} }[/tex] =0.67× G×[tex]\frac{m_{1} m_{2} }{R_{1} ^{2} }[/tex]
cancelleing like ternss from both sides we have
1/R₂² =0.67/R₁² or r₁²/R₂² =0.67 and R₁/R₂ = 0.8185
or R₂ = R₁/0.8185 or 1.222·R = (6.37×10⁶ m)/0.8185 = 7.783×10⁶ m
Hence at an altitude = 1.413×10⁶ m the weight of the object will reduce to 67% its earth-surface value
