Question

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move the satellite into a circular orbit with altitude 198 km? MJ (b) What is the change in the system's kinetic energy? MJ (c) What is the change in the system's potential energy?

Answer 1

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×$10^{24}$ 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×$10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=$\sqrt{\frac{Gmm}{R+h} }$         [(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 = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ($E_{i}$)  = $\frac{1}{2}$m$v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v=$\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h=$h_{f}$ = 198km = 198000 m

∴Final Energy($E_{f}$) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

        = $\frac{GMm}{2}$($\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$)

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$m[$v_{f} ^{2}$ - $v_{i} ^{2}$]

                                                          = $\frac{GMm}{2}$[$\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[$\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$]

                                                             = 2ΔE

                                                             = 968.907 MJ