Question

If a man weights 90 N on the surface of the earth,then the height above the surface of the earth of radius R, where the weight is 30 N is
A)0.73R
B)R/root 3
C)3R
D)root 3×R​

Answer 1

Answer:

W = mg

Weight is 1/3 times, so will gravitional acceleration .

Let height be x .

Gravitional acceleration is inversely proportional to square of distance from centre of earth

$\frac{(R + x) {}^{2} }{R {}^{2} } = \frac{3}{1} \\take \: \: root \\ R + x = \sqrt{3} R \\ \\ x = ( \sqrt{3} - 1)R = (1.7 - 1)R = 0.7R$

Answer 2

answer : option (1) 0.73R

weight of man at the surface of earth, $W_0$ = 90N

weight of man at h height above the surface of earth, W = 30N

using formula,

$W = \frac{W_0}{\left(1+\frac{h}{R}\right)^2}$

⇒ 30 = 90/(1 + h/R)²

⇒30/90 = 1/(1 + h/R)²

⇒3 = (1 + h/R)²

⇒√3 = 1 + h/R

⇒(√3 - 1) = h/R

⇒h = (√3 - 1)R = (1.73 - 1)R = 0.73R

hence, option (1) is correct choice.

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