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Chapter 8, Problem 73PQ

(a)

To determine

The height of the satellite’s orbit if its total energy were 500MJ greater.

(a)

Expert Solution
Check Mark

Answer to Problem 73PQ

The height of the satellite’s orbit if its total energy were 500MJ greater is 238km .

Explanation of Solution

Write the equation total energy for a satellite in a circular orbit.

  Etot=GmME2r

Here, Etot is the total energy of satellite in a circular orbit, G is universal gravitation constant, m is the mass of satellite, ME is the mass earth, r is the radius of circular orbit.

Write the expression for the energy difference of the satellite in two positions.

  ΔE=GmME2(1ri1rf)

Here, ri is the radius of initial orbit and rf is the radius of final orbit.

Rearrange above equation to get 1ri1rf.

  (1ri1rf)=2ΔEGmME                                                                                                   (I)

Initial position of the satellite is 125km from surface of earth. Let hf be the unknown height from the surface of earth.

Write the expression for ri and rf.

  ri=RE+125km                                                                                                       (II)

  rf=RE+hf                                                                                                             (III)

Here, RE is the radius of earth and hf is the final height of satellite.

Conclusion:

Radius of earth is 6370×103m and mass of earth is 5.98×1024kg.

Substitute 6370km for RE in equation (II) to get ri.

  ri=6370km+125km=6495km×1000m1km=6495×103m

Substitute 6370km for RE in equation (III) to get expression for rf.

  rf=6370km+hf

Substitute 6495×103m for ri, 6370km+hf for rf, 950kg for m, 5.98×1024kg for ME, 6.67×1011Nm2/kg2 for G and 500MJ for ΔE in equation (I) to get 1ri1rf.

  (1ri1rf)=2(500MJ×106J1MJ)(6.67×1011Nm2/kg2)(950kg)(5.98×1024kg)=2.64×109m1

Substitute 6495×103m for ri, 6370km+hf for rf in above equation to get hf.

  (16495×103m16370km+hf)=2.64×109m116370km+hf=16495×103m2.64×109m16370km+hf=6608311m×1km1000=6608.311km

Rearrange above equation to hf

  6370km+hf=6608.311kmhf=6608.311km6370km=238.311km238km

Therefore, the height of the satellite’s orbit if its total energy were 500MJ greater is 238km .

(b)

To determine

The difference in the system’s kinetic energy.

(b)

Expert Solution
Check Mark

Answer to Problem 73PQ

The kinetic energy of the system decreases by 500MJ.

Explanation of Solution

Total energy of an object in bound orbit is always negative. Therefore, for higher and lower orbits, the total energy of the satellite is negative. Its absolute value is equal to kinetic energy. For a circular orbit, total energy and potential energy can be related as E=12Ug where E is the total energy of the satellite and Ug is the potential energy of satellite. Taking changes in both side of above equation gives ΔE=ΔUg2.

For a circular orbit, kinetic energy and potential energy can be related as K=12Ug where K is the kinetic energy of the satellite and Ug is the potential energy of satellite. The equation indicates that kinetic energy is half that of potential energy. Taking change in both side of above equation gives ΔK=12ΔUg. Since ΔE=ΔUg2, change in kinetic energy takes the form ΔK=ΔE.

Therefore, change in kinetic energy is equal to negative change in total energy. In this problem, total energy of the satellite is increased by 500MJ. Thus kinetic energy decreases by 500MJ.

Conclusion:

Therefore, kinetic energy of the system decreases by 500MJ.

(c)

To determine

The difference in the system’s potential energy.

(c)

Expert Solution
Check Mark

Answer to Problem 73PQ

The potential energy of the system increases by 1000MJ.

Explanation of Solution

For an object in bound orbit, its total energy is always negative. Therefore, for higher and lower orbits, the total energy of the satellite is negative. Its absolute value is equal to kinetic energy.

For a circular orbit, total energy and potential energy can be related as E=12Ug where E is the total energy of the satellite and Ug is the potential energy of satellite. The equation indicates that potential energy is twice as that total energy. Taking changes in both side of above equation gives ΔE=ΔUg2. Thus, change in potential energy is equal to twice the change in total energy of satellite.

In this problem, total energy of the satellite is increased by 500MJ. Thus gravitational potential energy increases by 1000MJ.

Conclusion:

Therefore, potential energy of the system increases by 1000MJ.

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Chapter 8 Solutions

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term

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