Answer :
Answer:
Explanation:
Total energy of a satellite in an orbit , h height away
= - GMm /2 ( R + h )
When h = 380 km
Total energy of a satellite = [tex]\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}[/tex]
= - 13.25 x 10¹⁰ J
When h = 190 km
Total energy of a satellite =
[tex]\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}[/tex]
= - 13.63 x 10¹⁰ J
Diff
= 38 x 10⁸ J Energy will be required.
The energy required by an spacecraft to move into a higher orbit will be [tex]E=38\times10^{8}[/tex]J
What will be the energy required by the space craft to move into a higher orbit?
It is given that,
Mass of the spacecraft M=4500kg
Lower orbit =190lm
Higher orbit=380km
mass of earth [tex]M_{Earth}=5.97\times10^{24} kg[/tex]
Total energy of satellite in an orbit is given by formula
[tex]E=\dfrac{-GMm}{2(R+h)}[/tex]
Now for hieght h = 380 km
Energy of the satelite will be
[tex]E_{1} =\dfrac{6.67\times10^{-11}\times5.97\times10^{24}\times4500 }{2(6378+380)\times10^{3} }[/tex]
[tex]E_{1} = -13.25\times10^{10} J[/tex]
Now energy when the height of the satellite h=190km
[tex]E_{2} =\dfrac{6.67\times10^{-11}\times5.97\times10^{24}\times4500 }{2(6378+190)\times10^{3} }[/tex]
[tex]E_{2} =-13.63\times10^{10}[/tex]
Now the difference between the energies
[tex]E_{1}- E_{2}= 38\times10^{8} J[/tex]
Thus the energy required by an spacecraft to move into a higher orbit will be [tex]E=38\times10^{8}[/tex]J
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