A remote - sensing satellite of earth revolves in a circular orbit at a height of $0.25\times {{10}^{6}}$ m above the surface of earth. If earth's radius is $6.38\times {{10}^{6}}m$ and $g=9.8m{{s}^{-2}}$, then the orbital speed of the satellite is.
$\begin{align}
& \text{A}\text{. }6.67km{{s}^{-1}} \\
& \text{B}\text{. }7.76km{{s}^{-1}} \\
& \text{C}\text{. }8.56km{{s}^{-1}} \\
& \text{D}\text{. }9.13km{{s}^{-1}} \\
\end{align}$
Answer
Verified
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Hint: We know that the Orbital velocity is velocity required to put the satellite into its orbit around the earth. We know that in equilibrium the centripetal force is just provided by the gravitational pull of the earth. And from there we obtained the equation to find the orbital velocity.
Formula used:
$v=\sqrt{\dfrac{GM}{R+h}}$
Complete answer:
The orbital velocity of the satellite is
$v=\sqrt{\dfrac{GM}{R+h}}$
Where M= mass of earth and R = radius of earth.
So
$v=\sqrt{\dfrac{6.67\times {{10}^{-11}}\times \left( 6\times {{10}^{24}} \right)}{\left( 6.38\times {{10}^{6}} \right)+\left( 0.25\times {{10}^{6}} \right)}}=7.76km/s$
So, the correct answer is “Option B”.
Additional Information:
Natural satellite: A satellite created by nature is called a natural satellite. Moon is a natural satellite
Artificial satellite: A man made satellite is called an artificial satellite. Russian’s made the first artificial satellite called SPUTNIK-1.
Geostationary satellite: if a satellite is made to revolve from west to east with a period of revolution equal to 24 hrs in a circular orbit concentric and coplanar with the equatorial plane of earth, its relative velocity with respect to earth is zero. Such a satellite is called a geostationary satellite.
It should revolve at a height of nearly 36,000 km above the earth surface.
Its period of revolution around the earth should exactly be the same as that of the earth about its own axis i.e 24 hrs.
Note:
Angular velocity for a geostationary satellite through the equator is zero. Orbital velocity is velocity required to put the satellite into its orbit around the earth. Orbital velocity of a satellite is independent of the mass of the satellite decreases with the increase in the radius of the orbit and with the increase in the height of the satellite.
Formula used:
$v=\sqrt{\dfrac{GM}{R+h}}$
Complete answer:
The orbital velocity of the satellite is
$v=\sqrt{\dfrac{GM}{R+h}}$
Where M= mass of earth and R = radius of earth.
So
$v=\sqrt{\dfrac{6.67\times {{10}^{-11}}\times \left( 6\times {{10}^{24}} \right)}{\left( 6.38\times {{10}^{6}} \right)+\left( 0.25\times {{10}^{6}} \right)}}=7.76km/s$
So, the correct answer is “Option B”.
Additional Information:
Natural satellite: A satellite created by nature is called a natural satellite. Moon is a natural satellite
Artificial satellite: A man made satellite is called an artificial satellite. Russian’s made the first artificial satellite called SPUTNIK-1.
Geostationary satellite: if a satellite is made to revolve from west to east with a period of revolution equal to 24 hrs in a circular orbit concentric and coplanar with the equatorial plane of earth, its relative velocity with respect to earth is zero. Such a satellite is called a geostationary satellite.
It should revolve at a height of nearly 36,000 km above the earth surface.
Its period of revolution around the earth should exactly be the same as that of the earth about its own axis i.e 24 hrs.
Note:
Angular velocity for a geostationary satellite through the equator is zero. Orbital velocity is velocity required to put the satellite into its orbit around the earth. Orbital velocity of a satellite is independent of the mass of the satellite decreases with the increase in the radius of the orbit and with the increase in the height of the satellite.
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