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# The orbital speed for an Earth satellite near the surface of the earth is $7km{{s}^{-1}}$. If the radius of the orbit is $4$ times the radius of the Earth, the orbital speed would be.  Verified
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Hint: The orbital speed for an Earth satellite near the surface can be considered as the orbital speed for a satellite along the surface of the Earth. Therefore, the radius of the satellite is the same as the radius of the Earth. The ratio of orbital speed of a satellite at a particular distance away from the Earth to the orbital speed of a satellite along the surface of the Earth can be used to obtain the answer.

Formula used:
${{V}_{orbit}}=\sqrt{\dfrac{GM}{{{R}_{orbit}}}}$
where
${{V}_{orbit}}$ is the orbital speed of an object revolving around the Earth
${{R}_{orbit}}$ is the radius of the orbit in which the object revolves around the Earth
$G$ is the gravitational constant
$M$ is the mass of the satellite

Orbital speed is defined as the speed of an object revolving around a star or a planet in a definite orbit. Speed is usually calculated in $m{{s}^{-1}}$. But when we deal with the speeds of objects like satellites, planets or stars, orbital speed is usually measured in $km{{s}^{-1}}$. As mentioned in the formula, orbital speed is given by
${{V}_{orbit}}=\sqrt{\dfrac{GM}{{{R}_{orbit}}}}$
For a satellite orbiting near the surface of the Earth, the radius of the satellite is equal to the radius of the Earth and the orbital speed is calculated along the surface of the Earth. The above equation can be rewritten as
${{V}_{1}}=\sqrt{\dfrac{GM}{{{R}_{1}}}}$ (equation 1)
where
${{V}_{1}}$ is the orbital speed along the surface of the Earth
${{R}_{1}}$ is the radius of the Earth
Now, for a satellite orbiting in an orbit whose radius is equal to $4$ times that of the Earth at a certain distance from the Earth,
${{V}_{orbit}}=\sqrt{\dfrac{GM}{{{R}_{orbit}}}=}\sqrt{\dfrac{GM}{4{{R}_{1}}}}$ (equation 2)
Here,
${{R}_{orbit}}=4{{R}_{1}}$ (as given in the question)
Now, let us take the ratio of both orbital speeds i.e. the ratio of equation 1 and equation 2
$\dfrac{{{V}_{1}}}{{{V}_{orbit}}}=\dfrac{\sqrt{\dfrac{GM}{{{R}_{1}}}}}{\sqrt{\dfrac{GM}{4{{R}_{1}}}}}=\sqrt{\dfrac{GM}{{{R}_{1}}}}\times \sqrt{\dfrac{4{{R}_{1}}}{GM}}=\sqrt{4}=2$
We are provided that ${{V}_{1}}=7km{{s}^{-1}}$. Substituting this value in the above equation,
\begin{align} & \dfrac{{{V}_{1}}}{{{V}_{orbit}}}=\dfrac{7}{{{V}_{orbit}}}=2 \\ & {{V}_{orbit}}=\dfrac{7}{2}=3.5km{{s}^{-1}} \\ \end{align}

Note:
Mass of the satellite is the same in both the equations of orbital speed. Whether it moves along the surface of the Earth or at a distance away from the Earth, mass remains the same. Instead of calculating orbital speed by substituting values for each, students can directly go for the ratio method if comparisons are given in the question.