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Change in satellite's speed

  1. May 10, 2010 #1

    bon

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    1. The problem statement, all variables and given/known data

    1) A satellite is in circular orbit around the sun. Which requires the greater change in satellite's speed: to escape the solar system or fall into the sun?

    2. Relevant equations



    3. The attempt at a solution

    Ok so am i right in thinking that the velocity of such a satellite in circular orbit is root GM/(R+r) where R is radius of sun, r dist above surface of sun..

    so v escape is root 2GM (R+r) yes?

    So it requires a change by factor of root 2.. is that right?

    How do i calculate the change needed to fall into the sun?

    thanks
     
  2. jcsd
  3. May 10, 2010 #2

    D H

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    Re: Orbits

    There's a simple answer and a slightly more complex one, both of which are appropriate for an introductory physics class.

    The simple answer: How much would you have to change the satellite's velocity so the satellite falls straight into the Sun?

    The slightly more complex answer is by analogy. *Huge* rockets are needed to launch the Space Shuttle into orbit. A much smaller rocket is all that is needed to bring the Shuttle back to the Earth. Why is that?
     
  4. May 10, 2010 #3

    Andrew Mason

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    Re: Orbits

    Set out the equation for energy in terms of kinetic and potential energy in terms of r. What is KE in terms of r?

    Set r = Rsun and work out the speed
    Set [itex]r = \infty[/itex] and work out the speed.

    AM
     
  5. May 10, 2010 #4

    D H

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    Re: Orbits

    r here is the orbital altitude, that is the (circular) orbital radius less the radius of the Sun. It is a given.
     
  6. May 11, 2010 #5

    bon

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    Re: Orbits


    Hmm I guess because the Sun's gravitational field helps the small rocket come back down, whereas the huge rocket has to do work against the sun's gravity?

    So is my answer for vesc right?

    So would you have to make the velocity = 0 for it to fall straight to the sun? In which case it requires less to go off to infinity?
     
  7. May 11, 2010 #6

    Andrew Mason

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    Re: Orbits

    No. If the orbiting rocket loses all speed (v = 0), the change in kinetic energy is the same as the change when escaping (where v = 0). Assume it changes to an orbit that just intersects the sun's surface. It doesn't have to lose all of its speed to fall into the sun.

    You should work it out.

    AM
     
  8. May 11, 2010 #7

    D H

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    Re: Orbits

    No. The question is asking for the change in velocity needed to make an object in a circular orbit escape the solar system versus making it fall into the Sun. That change occurs at the initial circular orbit altitude. You are mixing things up here a bit, Andrew. The amount of energy needed to make the satellite escape is not the same as the amount of energy needed to make it fall straight into the Sun.

    Correct on both accounts. In post #2 I hinted, possibly too obliquely, at targeting the Sun's surface as opposed to targeting the center of the Sun (which is what canceling the satellite's velocity does). On the other account, bon, you need to work this out.

    -------------------

    Yes and no. You have the orbital velocity and escape velocity correct. However. you need to look at the change in velocity, [itex]||\vec v_{\text{new}} - \vec v_{\text{old}}||[/itex], not the ratio.
     
  9. May 11, 2010 #8

    Andrew Mason

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    Re: Orbits

    If you work it out, the magnitude of the change in KE is the same.

    Let r' = R + r

    In orbit:

    [tex]KE_{orbit} = \frac{1}{2}mv^2 = \frac{GMm}{2r'}[/tex]

    [tex]PE = -\frac{GMm}{r'}[/tex]

    Additional escape KE has to result in PE + KE =0 so:

    Additional KE has to equal:

    [tex]\Delta KE_{escape} = \frac{GMm}{2r'} = |\Delta KE_{orbit}|[/tex]

    The magnitude of the change in KE to escape is the same as the change in KE falling toward the centre of sun (ie. it loses all of its orbital KE). It can fall to the surface with less loss of orbital KE.

    AM
     
  10. May 11, 2010 #9

    D H

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    Re: Orbits

    Andrew, that is the change in energy from the perspective of an inertial observer in which the circularly-orbiting satellite has a velocity of sqrt(GMm/2(R+r)). That change in energy is a false metric, and it is not what the question asked. The question asked for change in velocity.

    The reason your change in kinetic energy is a false metric is because kinetic energy, and hence change in kinetic energy, is a frame-dependent quantity. Why pick a Sun-centered frame? A better (but still false) metric is the change in kinetic energy as observed from an inertial frame that instantaneously moving with the circularly orbiting satellite. Now the change in kinetic energy needed to escape is (1.5-√2)mv2 while the change in kinetic energy needed to dive straight into the Sun is 0.5mv2 (here v is the circular orbit velocity given by v2 = GM/2(R+r)). From the perspective of this frame, it takes quite a bit less energy to escape the solar system than it does to dive straight into the Sun.

    In the limit of a rocket with an infinite exhaust velocity, the above would reflect the amount of energy needed to make the satellite escape versus dive into the Sun. This is still a false metric, however. Real rockets do not have an infinite exhaust velocity. The governing equation for rockets is the ideal rocket equation. The variable in that equation is Δv, the change in velocity. The question asked for the change in velocity precisely because this is the driving factor in determining the amount of fuel needed by a space vehicle to accomplish its mission.
     
  11. May 12, 2010 #10

    Andrew Mason

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    Re: Orbits

    But you work out the change in velocity from the change in kinetic energy. The changes in kinetic energy are the same as between: 1. losing all orbital kinetic energy and falling to the centre of the sun and 2. doubling orbital kinetic energy to escape (ie v=0 at infinity). From that, one can compare the third possibility 3. the (obviously smaller) change in kinetic energy required to fall just onto the surface of the sun. I did not say that this represents the same change in velocity. But from the changes in KE one can see which of the three represents the greatest change in velocity.

    So is velocity.
    Because it is apparent from the question that the sun-centred frame is the one in which the speed is being measured.

    AM
     
  12. May 12, 2010 #11

    D H

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    Re: Orbits

    So? That does not measure the amount of energy that needs to be expended by the vehicle to change the orbit.

    You are arguing about something that I happen to know a lot about, Andrew. Orbital mechanics and spacecraft dynamics is what I do for a living. It always takes a lot more energy (read as more fuel) to make a vehicle that is in a circular orbit about the Sun dive straight into the Sun than it does to make that vehicle escape the solar system. Except for vehicles orbiting very close to the Sun (between one and ten solar radii; I don't want to give the precise number as that would be answering this homework problem), it takes a lot more energy to make such vehicles just skim the Sun's surface that it does to make the vehicle escape the solar system. For a vehicle launched from the Earth, it takes a lot more energy to send a vehicle to Mercury from Earth (let alone diving into the Sun) than it does to make the vehicle escape the solar system.

    Largely because of the excessive Δv costs, the world's space agencies have sent only two vehicles to Mercury. The last such mission, MESSENGER, used six gravity assists to cut down on those Δv costs: One from the Earth a year after launch, two from Venus, and three from Mercury itself. Without those gravity assists a mission to Mercury would be well beyond the reach of current propulsion technology.

    Δv is frame independent in a Newtonian world, and that is the world in which the puny spacecraft invented by mankind live.
     
    Last edited: May 12, 2010
  13. May 12, 2010 #12

    Andrew Mason

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    Re: Orbits

    I did not suggest that it did. The amount of energy expended by the rocket bears little relationship to the change in kinetic energy of the rocket - a rocket launch from earth shows that very well. I was only interested in the change in KE of the rocket, from which one can calculate the needed change in v.

    And arguing is something I know a lot about. Arguing is what I do for a living.:wink:

    My point was that the change in kinetic energy of the rocket is the same and you can work out the change in velocities from that. You have pointed out that the fact that the changes in KE of the rocket are the same does not mean that the same amount of rocket fuel is needed to effect such changes, and I quite agree.

    These are all very good points. The [itex]\Delta v [/itex] is important parameter that determines the amount of rocket energy needed, rather than the change in KE of the rocket. But in this question, it seems to me that you have to work out the needed change in KE of the rocket in order to determine the required [itex]\Delta v [/itex]. That is really my only point. How else would you determine the [itex]\Delta v [/itex]?

    AM
     
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