Respuesta :
Answer:
a) The Energy added should be 484.438 MJ
b) The Kinetic Energy change is -484.438 MJ
c) The Potential Energy change is 968.907 MJ
Explanation:
Let 'm' be the mass of the satellite , 'M'(6×[tex]10^{24}[/tex] be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×[tex]10^{-11}[/tex] N/m) be the universal constant of gravitation.
We know that the orbital velocity(v) for a satellite -
v=[tex]\sqrt{\frac{Gmm}{R+h} }[/tex] [(R+h) is the distance of the satellite from the center of the earth ]
Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)
For initial conditions ,
h = [tex]h_{i}[/tex] = 98 km = 98000 m
∴Initial Energy ([tex]E_{i}[/tex]) = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] + [tex]\frac{-GMm}{(R+h_{i} )}[/tex]
Substituting v=[tex]\sqrt{\frac{GMm}{R+h_{i} } }[/tex] in the above equation and simplifying we get,
[tex]E_{i}[/tex] = [tex]\frac{-GMm}{2(R+h_{i}) }[/tex]
Similarly for final condition,
h=[tex]h_{f}[/tex] = 198km = 198000 m
∴Final Energy([tex]E_{f}[/tex]) = [tex]\frac{-GMm}{2(R+h_{f}) }[/tex]
a) The energy that should be added should be the difference in the energy of initial and final states -
∴ ΔE = [tex]E_{f}[/tex] - [tex]E_{i}[/tex]
= [tex]\frac{GMm}{2}[/tex]([tex]\frac{1}{R+h_{i} }[/tex] - [tex]\frac{1}{R+h_{f} }[/tex])
Substituting ,
M = 6 × [tex]10^{24}[/tex] kg
m = 1036 kg
G = 6.67 × [tex]10^{-11}[/tex]
R = 6400000 m
[tex]h_{i}[/tex] = 98000 m
[tex]h_{f}[/tex] = 198000 m
We get ,
ΔE = 484.438 MJ
b) Change in Kinetic Energy (ΔKE) = [tex]\frac{1}{2}[/tex]m[[tex]v_{f} ^{2}[/tex] - [tex]v_{i} ^{2}[/tex]]
= [tex]\frac{GMm}{2}[/tex][[tex]\frac{1} {R+h_{f} }[/tex] - [tex]\frac{1} {R+h_{i} }[/tex]]
= -ΔE
= - 484.438 MJ
c) Change in Potential Energy (ΔPE) = GMm[[tex]\frac{1}{R+h_{i} }[/tex] - [tex]\frac{1}{R+h_{f} }[/tex]]
= 2ΔE
= 968.907 MJ
