Answer
414.9k+ views
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.
Given,
\[\begin{align}
& R=6370km \\
& g=9.8m{{s}^{-2}} \\
\end{align}\]
Let us derive the equation for height ‘h’ from the above formula of time period.
i.e.,
\[\begin{align}
& {{T}^{2}}=4{{\pi }^{2}}\dfrac{{{\left( R+h \right)}^{2}}}{g{{R}^{2}}} \\
& \Rightarrow \dfrac{g{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{2}} \\
\end{align}\]
The time period should be in seconds.
\[\begin{align}
& T=90minutes \\
& T=90\times 60\times 60 \\
& \Rightarrow T=5400\sec \\
\end{align}\]
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 \[h=280km\].
Note:
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.
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.
Given,
\[\begin{align}
& R=6370km \\
& g=9.8m{{s}^{-2}} \\
\end{align}\]
Let us derive the equation for height ‘h’ from the above formula of time period.
i.e.,
\[\begin{align}
& {{T}^{2}}=4{{\pi }^{2}}\dfrac{{{\left( R+h \right)}^{2}}}{g{{R}^{2}}} \\
& \Rightarrow \dfrac{g{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{2}} \\
\end{align}\]
The time period should be in seconds.
\[\begin{align}
& T=90minutes \\
& T=90\times 60\times 60 \\
& \Rightarrow T=5400\sec \\
\end{align}\]
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 \[h=280km\].
Note:
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)