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A satellite which is at a height h from the surface completes one revolution of the earth in 90 minutes. If the radius of the earth is 6370km and acceleration due to gravity 9.8\[m{{s}^{-2}}\]. Then find h.

Last updated date: 13th Jun 2024
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Hint: We have to find the height at which the satellite is orbiting by finding the radius of the orbit by substituting in the equation for the time period of revolution of the satellite around the earth. From that we can find the height by substituting in the formulae.

Complete answer:
We can find the height at which the satellite is orbiting using the equation for the time period of revolution.
From the laws of gravity and energy conservations we have derived the equation for time period as –
Time period of satellite,
\[T=2\pi \sqrt{\dfrac{{{\left( R+h \right)}^{2}}}{g{{R}^{2}}}}\]
where, R is the radius of the earth, h is the height of the orbit and g is the acceleration due to gravity.
  & R=6370km \\
 & g=9.8m{{s}^{-2}} \\
Let us derive the equation for height ‘h’ from the above formula of time period.
  & {{T}^{2}}=4{{\pi }^{2}}\dfrac{{{\left( R+h \right)}^{2}}}{g{{R}^{2}}} \\
 & \Rightarrow \dfrac{g{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{2}} \\
The time period should be in seconds.
  & T=90minutes \\
 & T=90\times 60\times 60 \\
 & \Rightarrow T=5400\sec \\
Substituting the values for the radius of earth, ‘r’, height of the orbit ‘h’ and acceleration due to the gravity ‘g’, we get,
\[\Rightarrow \dfrac{9.8\times {{6370}^{2}}\times {{5400}^{2}}}{4\times {{3.14}^{2}}}={{(6370+h)}^{2}}\]
On finding the solution, we will get
 \[\Rightarrow h=280km\]

The height of the orbit at which the satellite revolves with a time period 90minutes is at \[h=280km\].

The time period of revolution is the time taken by the satellite to complete one full cycle around the earth. As the satellite moves from outer and outer to the space, this time period increases. It is clear that as it moves away from the Earth the height increases and the radius of revolution increases.