Question

Find the height over the earth's surface at which the weight of a body becomes half of its value at the surface

Answer 1

Answer:

Explanation:

Given, weight of the body becomes half.

Let that height be h.

  mg/2  =  GMm/(R+h)²

and, g = GM/(R)²  ⇒  1/2  =  1/(1 + h/R)²

hence, h  =  (√2 - 1) R

Multiplying numerator and denominator with (√2 + 1)

h  =  R / (√2 + 1)

Hope it helps .

ありがとうございました

Arigatōgozaimashita

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Answer 2

Answer:

$\text { The height over the earth's surface is } h=(\sqrt{2}-1) R \text {. }$

Explanation:

$Tip$

Newton's law of Gravitation: $F=\frac{G M_{1} M_{2}}{r^{2}}$

This is also known as the gravitational constant.

Explanation

Given:

$F=2 F^{\prime}$

To Find:

The height over the earth's surface.

Calculate the height over the earth's surface.

Calculation: The weight on earth is:

$F=\frac{G M m}{R^{2}}$

Weight at height $h$

$F^{\prime}=\frac{G M m}{(R+h)^{2}}$

Use both the above equations, and calculate the height.

Calculation:

$\Rightarrow 2 \frac{G M m}{(R+h)^{2}}=\frac{G M m}{R^{2}}$

$\begin{aligned}&\Rightarrow(R+h)^{2}=2 R^{2} \\&\Rightarrow h=(\sqrt{2}-1) R\end{aligned}$

$\text { Thus, the height over the earth's surface is } h=(\sqrt{2}-1) R \text {. }$