Question 302: Gravitation

Question

A satellite is revolving around a planet in a circular orbit close to its surface. Let ' $\rho$ ' be the mean density and ' $R$ ' be the radius of the planet. Then the period of the satellite is ( $\mathrm{G}=$ universal constant of gravitation)

Accepted Answer

From Kepler's third law,
$$\begin{aligned}
\mathrm{T}^2 & =\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}} \
	herefore \quad \mathrm{~T} & =2 \pi \sqrt{\frac{\mathrm{r}^3}{\mathrm{GM}}}
\end{aligned}$$
As the satellite is very close to the planet, $r=R$.
$$	herefore \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}}\quad	ext{.... (i)}$$
We know, Mass $=$ Volume $	imes$ Density $(ho)$
$$=\frac{4}{3} \pi \mathrm{R}^3 	imes ho\quad	ext{.... (ii)}$$
Putting (ii) into (i)
$$T=2 \pi \sqrt{\frac{R^3}{G 	imes \frac{4}{3} \pi R^3 ho}}=\sqrt{\frac{3 \pi}{ho G}}$$

Answer

$\sqrt{\frac{4 \pi}{\rho G}}$

Answer

$\sqrt{\frac{\pi}{\rho G}}$

Answer

$\sqrt{\frac{3 \pi}{\rho G}}$

Answer

$\sqrt{\frac{2 \pi}{\rho G}}$

Gravitation

A satellite is revolving around a planet in a circular orbit close to its surface. Let ' \rho ' be the mean density and ' R ' be the radius of the planet. Then the period of the satellite is ( \mathrm{G}= universal constant of gravitation)

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