MHT CET · Physics · Gravitation
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
- A \(\frac{\mathrm{W}}{2}\)
- B \(\frac{2 \mathrm{~W}}{3}\)
- C \(\frac{4 \mathrm{~W}}{9}\)
- D \(\frac{8 W}{27}\)
Answer & Solution
Correct Answer
(C) \(\frac{4 \mathrm{~W}}{9}\)
Step-by-step Solution
Detailed explanation
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, \(r=\left(R+\frac{R}{2}\right)\)
\(\begin{aligned} & g^{\prime}=\left(\frac{\mathrm{GM}}{\mathrm{R}^2}\right)\left(\frac{4}{9}\right)=\frac{4}{9} \mathrm{~g}_0 \\ & \therefore \mathrm{W}^{\prime}=\mathrm{mg}=\frac{4}{9}\left(\mathrm{mg}_0\right)=\frac{4}{9} \mathrm{~W}\end{aligned}\)
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