Question

An object moves from the earth’s surface to the surface of the moon. The acceleration due to gravity on the earth’s surface is$10m/s^2$. Considering the acceleration due to gravity on the moon to be$1/6thtimes$that of earth. If$R$ be the earth’s radius and its weight be $W$and the distance between the earth and the moon is$D$. The correct variation of the weight$W’$versus distance $d$ for a body when it moves from the earth to the moon is

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The correct option is C

Step 1. Given data :

At the earths surface , the weight of the body is ,$W=mg$

At the moon , the distance of the body from the earth is ,$d=D$

At the moons surface , the value of acceleration due to gravity is$g’=g/6\Rightarrow W’=W/6$

Step 2. Find the acceleration due to gravity at height d above earth’s surface:

$g\text{'}=\frac{g{R}^2}{{(R+d)}^2}-\left(i\right)$

For$d=0,\Rightarrow g’=gorW’=W$

$$\begin{gathered} atd=R, \\ g’=\frac{g{R}^2}{4R^2} \\ g’=\frac{g}{4} \\ Or,W’=\frac{mg}{4} \\ W’=\frac{W}{4} \end{gathered}$$

At distance$d=D,$ when the body reaches the surface of the moon

$$\begin{gathered} W’=\frac{mg}{6} \\ W’=\frac{W}{6} \end{gathered}$$

Since $Eq.\left(i\right)$ suggests that variation of$g$ with distance $\left(d\right)$ is non-linear, hence the graph of

$W\prime vsd$will be non-linear as well. Also,$d\uparrow ,W’\downarrow$

Thus, the correct variation is represented by the following graph

Hence, (C) is the correct answer.