$A$ body is moving in a low circular orbit about a planet of mass $M$ and radius $R$. The radius of the orbit can be taken to be $R$ itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is

  • A
    $1$
  • B
    $2$
  • C
    $\frac{1}{\sqrt{2}}$
  • D
    $\sqrt{2}$

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Similar Questions

The radius of the orbit of a geostationary satellite is (mean radius of earth is $R$,angular velocity about own axis is $\omega$,and acceleration due to gravity on earth's surface is $g$).

$A$ satellite of mass $m$ is revolving close to the surface of a planet of density $d$ with time period $T$. The value of the universal gravitational constant $G$ in terms of $d$ and $T$ is given by:

If the gravitational constant $G$ is decreasing with time,what will remain unchanged for a satellite orbiting around the Earth?

$A$ 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$

An artificial satellite is moving around the Earth in a circular orbit with a speed equal to one-fourth the escape speed of a body from the surface of the Earth. The height of the satellite above the Earth's surface is ............ ($R$ is the radius of the Earth). (in $R$)

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