An object at the surface of earth weighs $90 N$ . Its weight at a distance of $3 R$ (where R is the radius of earth) from the center of earth is $\dots \dots \dots$

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Solution

Concept of weight
According to Newton's law of gravitation,
Weight $W = m \times g$ ; where $m = massofobject$ and $g = accelerationduetogravity$
Also, $g = \frac{G M}{R^{2}}$ , where G=gravitational constant, M=mass of earth, R=radius of earth.
Hence, $W = m \times \frac{G M}{R^{2}}$
Weight is inversely proportional to the radius R
Value of acceleration due to gravity, $g = 9.8 {ms}^{- 2}$
Given data and calculation:
Weight of objects at the surface of the earth $W_{1} = 90 N$
Let $W_{2}$ be the weight at a distance of 3 R from the center.
So, $\begin{array}{rcl} \frac{W_{1}}{W_{2}} & = & \frac{{(3 R)}^{2}}{R^{2}} \\ = & 9 \end{array}$
Weight at a distance 3 R $= \frac{90}{9} = 10 N$
Thus, weight at a distance of $3 R$ (where R is the radius of the earth) from the center of the earth is $10 N$

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An object at the surface of earth weighs 90N. Its weight at a distance of 3R (where R is the radius of earth) from the center of earth is ………

An object at the surface of earth weighs $90 N$ . Its weight at a distance of $3 R$ (where R is the radius of earth) from the center of earth is $\dots \dots \dots$