Question

At what altitude h above the north pole is the weight of an object reduced to 67% of its earth-surface value? Assume a spherical earth of radius R and express h in terms of R.

Answer 1

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$\frac{m_{1} m_{2} }{R^{2} }$  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$\frac{m_{1} m_{2} }{R_{2} ^{2} }$ =0.67× G×$\frac{m_{1} m_{2} }{R_{1} ^{2} }$

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