Question

A body weighs 81 N on the surface of the earth. How much will it weigh when taken to a height equal to half of radius of earth.

Answer 1

Step 1: Given

$$\begin{gathered} WeightofobjectonEarth,W_E=81N \\ Dis\tan cewhenobjectistakentoaheight,R_{new}=R+\frac{R}{2}=\frac{3R}{2} \\ where, \\ R\to RadiusoftheEarth \\ Weightatthenewheight,W_{new}=? \end{gathered}$$

Step 2: Formula used

Writing down the Weight on earth's surface in terms of mass and acceleration due to gravity

$$\begin{gathered} W_E=m\times g_E \\ where, \\ m\to massoftheobject \\ {g}_E\to accelerationduetogravityonthesurfaceoftheEarth \\ So, \\ {W}_E=m\times \frac{GM}{R^2} \\ where, \\ M\to Massoftheearth \end{gathered}$$

Step 3: Calculation

Writing down the expression of weight at the new height

$W_{new}=m\times \frac{GM}{{\left({\displaystyle \frac{3R}{2}}\right)}^2}$

Comparing both the expressions and substituting the value of W_E

$$\begin{gathered} W_{new}=m\times \frac{GM}{{\left({\displaystyle \frac{3R}{2}}\right)}^2} \\ {W}_E=m\times \frac{GM}{R^2} \\ \Rightarrow \frac{W_{new}}{W_E}=\frac{m\times \frac{GM}{{\left({\displaystyle \frac{3R}{2}}\right)}^2}}{m\times \frac{GM}{R^2}}=\frac{4}{9} \\ \Rightarrow \frac{W_{new}}{81}=\frac{4}{9} \\ \Rightarrow W_{new}=36N \end{gathered}$$

Hence, the weight of the object at a height equal to half of the radius of the earth will be $36N$.