Work in moving a satellite's orbit

  • Thread starter Thread starter gboff21
  • Start date Start date
  • Tags Tags
    Orbit Work
Click For Summary
SUMMARY

The discussion focuses on calculating the work required to move a 2580-kg spacecraft from a circular orbit 1580 km above Mars to a higher orbit 4190 km above the surface. The key equations used include the change in potential energy, ΔU, and the change in kinetic energy, ΔK, derived from gravitational principles. The final calculation yields a total energy change of approximately 3×1011 J. A critical correction was made regarding the factor of 3 in the energy equation, clarifying that ΔK is related to ΔU through a factor of 1/2.

PREREQUISITES
  • Understanding of gravitational potential energy (ΔU) and kinetic energy (ΔK)
  • Familiarity with the gravitational constant (G) and its application in orbital mechanics
  • Knowledge of circular motion dynamics and energy conservation principles
  • Basic algebra for manipulating equations involving physical constants and variables
NEXT STEPS
  • Study the derivation of gravitational potential energy in orbital mechanics
  • Learn about the implications of energy conservation in satellite motion
  • Explore the effects of altitude changes on orbital velocity and energy
  • Investigate the role of the gravitational constant (G) in different celestial contexts
USEFUL FOR

Aerospace engineers, physics students, and anyone involved in orbital mechanics or satellite design will benefit from this discussion.

gboff21
Messages
48
Reaction score
0

Homework Statement


A 2580-kg spacecraft is in a circular orbit a distance 1580km above the surface of Mars.
How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4190km above the surface?

Homework Equations


\DeltaU=GMsMm(1/r_2-1/r_1)
\DeltaK=1/2GMsMm(1/r_2-1/r_1)

\DeltaK derived from \frac{mv<sup>2</sup>}{r}=\frac{GM<sub>s</sub>M<sub>m</sub>}{r<sup>2</sup>}

The Attempt at a Solution


Adding the energies
\DeltaE=\DeltaU+\DeltaK=3/2GMsMm(1/r_2-1/r_1)

M_s=2580kg
M_m=6.4191×1023kg
r_m=3397 km
r_1=1580km
r_2=4190km
substituting gives:

\DeltaU+\DeltaK=3/2G*2580*6.4191×1023*(1/(4190×10^3-339710^3)-1/(1580×10^3-3397×10^3)=3×10^11 J

Is this right? Or am I missing something?
 
Physics news on Phys.org
gboff21 said:

The Attempt at a Solution


Adding the energies
\DeltaE=\DeltaU+\DeltaK=3/2GMsMm(1/r_2-1/r_1)

I don't see where this 3 comes from You can easily show that, for circular satellite motion,

K = - (1/2)U

So ΔK = -(1/2)ΔU, therefore

ΔK + ΔU = + (1/2)ΔU.
 
Ah I forgot that U is negative. So forget the 3 thanks
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
Replies
5
Views
3K
Replies
1
Views
2K
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
13K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 6 ·
Replies
6
Views
2K