$A$ 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 ($G=$ universal constant of gravitation).

  • A
    $\sqrt{\frac{4 \pi}{\rho G}}$
  • B
    $\sqrt{\frac{\pi}{\rho G}}$
  • C
    $\sqrt{\frac{3 \pi}{\rho G}}$
  • D
    $\sqrt{\frac{2 \pi}{\rho G}}$

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$A$ geostationary satellite is orbiting the earth at a height of $6R$ from the earth's surface ($R$ is the earth's radius). What is the period of rotation of another satellite at a height of $2.5R$ from the earth's surface?

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$A$ planet of mass $M$ has two natural satellites with masses $m_1$ and $m_2$. The radii of their circular orbits are $R_1$ and $R_2$ respectively. Ignore the gravitational force between the satellites. Define $v_1, L_1, K_1$ and $T_1$ to be,respectively,the orbital speed,angular momentum,kinetic energy,and time period of revolution of satellite $1$; and $v_2, L_2, K_2$ and $T_2$ to be the corresponding quantities of satellite $2$. Given $m_1/m_2 = 2$ and $R_1/R_2 = 1/4$,match the ratios in List-$I$ to the numbers in List-$II$.
List-$I$List-$II$
$P. \frac{v_1}{v_2}$$1. \frac{1}{8}$
$Q. \frac{L_1}{L_2}$$2. 1$
$R. \frac{K_1}{K_2}$$3. 2$
$S. \frac{T_1}{T_2}$$4. 8$

The orbit of a geostationary satellite is circular. The time period of the satellite depends on $(i)$ mass of the satellite,$(ii)$ mass of the earth,$(iii)$ radius of the orbit,and $(iv)$ height of the satellite from the surface of the earth.

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