A 970-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How muc

Question

A 970-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 208 km? MegaJoules! (MJ)

1E9

Incorrect: Your answer is incorrect.

How is the total energy of an object in circular orbit related to the potential energy? MegaJoules! (MJ)

(b) What is the change in the system's kinetic energy?

-1E9

Incorrect: Your answer is incorrect.

You appear to have calculated the change in kinetic energy correctly using your incorrect result from part (a). Megajoules! (MJ)

(c) What is the change in the system's potential energy?

2E^9

Incorrect: Your answer is incorrect.
Your answer cannot be understood or graded. More Information MJ

Explanation / Answer

for circular path,

Gravitational force = m x centripetal acc

G M m / r^2 = m v^2 / r

m v^2 = G M m / r

total energy = PE + KE

= - G M m / r + m v^2 / 2

= - G M m / r + G M m /2r

= - G M m / 2 r

for h = 98 km , r = 6371 + 98 = 6469 km = 6469 x 10^3 m

h = 208 km. r = 6579 x 10^3 m

Ei= - (6.67 x 10^-11)(5.972 x 10^24)(970) / 2(6469 x 10^3)

= - 2.9864 x 10^10 J

Ef = -(6.67 x 10^-11)(5.972 x 10^24)(970) / 2(6579 x 10^3)

Ef = - 2.9365 x 10^10J

energy required = Ef - Ei = 5 x 10^8 J

1 MJ = 10^6 J

in MJ = 500 MJ ..........Ans(a)

(b) KEi = - Ei =2.9864 x 10^10 J

KEf= - Ef =  2.9365 x 10^10J

change in KE = KEf - KEi = - 5 x 10^8J

Ans: - 500 MJ

(c) PE = 2Ei and PEf = 2Ef

change in potential energy = PEf - PEi = 2(Ef - Ei)

= 1000 MJ

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