Search any question & find its solution
Question:
Answered & Verified by Expert
A body weighs ' $W$ ' newton on the surface of the earth its weight at a height equal to half the radius of the earth, will be
Options:
Solution:
1529 Upvotes
Verified Answer
The correct answer is:
$\frac{4 \mathrm{~W}}{9}$
The acceleration due to gravity at a distance $r$ from center of the earth is:
$\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{r}^2}$
For, $r=\left(R+\frac{R}{2}\right)$
$\begin{aligned} & g^{\prime}=\left(\frac{\mathrm{GM}}{\mathrm{R}^2}\right)\left(\frac{4}{9}\right)=\frac{4}{9} \mathrm{~g}_0 \\ & \therefore \mathrm{W}^{\prime}=\mathrm{mg}=\frac{4}{9}\left(\mathrm{mg}_0\right)=\frac{4}{9} \mathrm{~W}\end{aligned}$
$\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{r}^2}$
For, $r=\left(R+\frac{R}{2}\right)$
$\begin{aligned} & g^{\prime}=\left(\frac{\mathrm{GM}}{\mathrm{R}^2}\right)\left(\frac{4}{9}\right)=\frac{4}{9} \mathrm{~g}_0 \\ & \therefore \mathrm{W}^{\prime}=\mathrm{mg}=\frac{4}{9}\left(\mathrm{mg}_0\right)=\frac{4}{9} \mathrm{~W}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.