Conservation of Energy in Satellite Motion

In summary, the smaller piece of mass m falls directly toward the Earth after the explosion. The speed v of the larger piece immediately after the explosion is given by the equation v_f=\sqrt{\dfrac{5GM_E}{4(R_E+h)}}.
  • #1
38
0

Homework Statement


A satellite moves around the Earth in a circular orbit of radius r.
(a) What is the speed vi of the satellite?
(b) Suddenly, an explosion breaks the satellite into two pieces, with masses m and 4m. Immediately after the explosion, the smaller piece of mass m is stationary with respect to the Earth and falls directly toward the Earth. What is the speed v of the larger piece immediately after the explosion?
(c) Because of the increase in its speed, this larger piece now moves in a new elliptical orbit. Find its distance away from the center of the Earth when it reaches the the other end of the ellipse

Homework Equations


ΔE = ΔK + ΔU = 0
ΔƩmv = 0

The Attempt at a Solution


The apparently correct solution involves using the conservation of momentum. I tried to use the conservation of kinetic energy. My reasoning is that if the smaller piece of mass m is basically at rest after the explosion, it no longer has kinetic energy, thus all of the final kinetic energy is in the moving piece of mass 4m. I assume that the original mass is 5m. Equating the original kinetic energy of the object before the explosion to the kinetic energy of the smaller object, I got the following result:

[tex] K_i=\dfrac{1}{2}m_iv_i^2=\dfrac{5}{2}m\left(\dfrac{GM_E}{R_E+h}\right) [/tex]
[tex] K_f=2mv_f^2 [/tex]
[tex] \dfrac{5}{2}m(\dfrac{GM_E}{R_E+h})=2mv_f^2 [/tex]
[tex] v_f=\sqrt{\dfrac{5GM_E}{4(R_E+h)}} [/tex]

This is apparently wrong, but I'm not sure why. One thing that troubles me is that I do not account for potential gravitational energy. Keep in mind the correct answer, according to the solution's manual, is:

[tex] v_f=\dfrac{5}{4}\sqrt{\dfrac{GM_E}{R_E+h}} [/tex]

Also I do not know how to begin part (c). Intuitively I'm thinking that if the satellite is on the "other side" of an elliptical orbit, it's distance from the surface of the Earth would be the same as it's distance at it's starting point...suggestions?
 
Physics news on Phys.org
  • #2
Bennigan88 said:
The apparently correct solution involves using the conservation of momentum.

That would be correct - because you do not know whether the explosion transferred any of its energy to the remnants. It very well could.

The good thing is that after you get the momentum of the larger remnant, you can compute its kinetic energy as well.
 
  • #3
If the explosion was from fuel on board, then the system would be closed, and energy should be conserved. If the explosion was from a missile, for instance, then the system wouldn't be closed and I wouldn't be able to calculate the momentum of the larger piece of the satellite without first knowing the momentum of the missile at the instant of collision. Yikes!

But I still don't know how to approach part (c)
 
  • #4
Bennigan88 said:
If the explosion was from fuel on board, then the system would be closed, and energy should be conserved. If the explosion was from a missile, for instance, then the system wouldn't be closed and I wouldn't be able to calculate the momentum of the larger piece of the satellite without first knowing the momentum of the missile at the instant of collision. Yikes!

Total energy would be conserved. But we are talking about the mechanical energy here. Explosions usually convert huge amounts of chemical energy into mechanical.

But I still don't know how to approach part (c)

I think you will need conservation of energy and angular momentum. Or, if you have already studied motion in a central force field or the two-body problem, you could their results.
 
  • #5
For part (c), you should be able to calculate the orbital total mechanical energy ##\xi## for the object in question. ##\xi## is related to the major axis of an elliptical orbit (how?). Since the pre-explosion condition was a circular orbit, given the conditions of the explosion event it seems that at that instant the object in question will be at the perigee point of its new orbit (why?).

Note that elliptical orbits do NOT center the ellipse on the orbited body -- the primary is located at one focus of the ellipse. Thus the perigee and apogee are not at the same distance from the Earth!
 
  • #6
gneill said:
For part (c), you should be able to calculate the orbital total mechanical energy ##\xi## for the object in question. ##\xi## is related to the major axis of an elliptical orbit (how?). Since the pre-explosion condition was a circular orbit, given the conditions of the explosion event it seems that at that instant the object in question will be at the perigee point of its new orbit (why?).

Note that elliptical orbits do NOT center the ellipse on the orbited body -- the primary is located at one focus of the ellipse. Thus the perigee and apogee are not at the same distance from the Earth!

Ah, okay I see what you are saying at the perigee. I wasn't making the connection between how the increase in velocity affected the point on the ellipse (the new orbit) that the planet would now be at. I'll take a closer look at the relationship between the total mechanical energy and the semi-major axis a.

[tex] E = -\dfrac{GMm}{2a} [/tex]

I'll give this a shot and see if it comes together.
 
  • #7
##\mu = G(M + m) \approx GM## since ##M >> m##.
 

1. What is conservation of energy in satellite motion?

Conservation of energy in satellite motion refers to the principle that the total energy of a satellite in orbit around a planet remains constant. This means that the satellite's kinetic energy and potential energy are always in balance, and any changes in one type of energy must be offset by an equal and opposite change in the other.

2. How does conservation of energy relate to satellite orbits?

Conservation of energy is crucial for understanding the stability and maintenance of satellite orbits. Satellites in orbit are constantly moving, and conservation of energy ensures that they do not lose or gain too much energy, which could cause them to spiral out of orbit or crash into the planet.

3. What factors affect the conservation of energy in satellite motion?

The conservation of energy in satellite motion is influenced by factors such as the mass of the satellite, the distance from the planet, and the speed at which it is orbiting. These factors can all impact the balance between kinetic and potential energy in the satellite's orbit.

4. How is conservation of energy calculated in satellite motion?

Conservation of energy in satellite motion is calculated using the formula E = K + U, where E represents the total energy, K represents the kinetic energy, and U represents the potential energy. By plugging in the appropriate values for each type of energy, the total energy of the satellite can be determined.

5. What are some real-world applications of conservation of energy in satellite motion?

Conservation of energy in satellite motion is essential for the proper functioning of many technologies, including GPS systems, weather forecasting, and communication networks. By understanding and applying this principle, scientists and engineers can ensure the stability and longevity of satellites in orbit around the Earth.

Suggested for: Conservation of Energy in Satellite Motion

Back
Top