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Change in Gravitational Potential

  1. Jul 27, 2010 #1
    The Earth may be assumed to be an isolated sphere of radius 6.4 × 103 km with its mass of 6.0 × 1024 kg concentrated at its centre. An object is projected vertically from the surface of the Earth so that it reaches an altitude of 1.3 × 104km.

    Calculate, for this object:

    (i) the change in gravitational potential

    (ii) the speed of projection from the Earth’s surface, assuming air resistance is
    negligible.


    My Attempt

    i) I applied the formula

    φ = - GM/radius of earth + altitude of object
    = GM (1/radius of earth - 1/altitude of object)
    = 3.17 x 10^10 J/kg

    Don't know what I'm doing wrong here, this is not the correct answer.

    The marking scheme of this past paper states the following solution:

    change = 6.67 x 10^-11 x 6.0 x 10^24 x ({6.4 x 10^6}-1 - {1.94 x 10^7}-1)
    change = 4.19 x 107 J/kg

    I am not sure how they calculated the bold part?


    ii)

    I used the Escape Velocity Formula as follows:

    Vesc = squareroot{[2GM/r]} or squareroot[2gr]
    Vesc = 3.54 x 10^5

    Again this is not correct answer. So when do we use this escape velocity formula exactly if not in this case?

    Again the marking scheme has the following suggested solution:

    ½mv2 = mdeltaφ
    v2 = 2 x 4.19 x 107 = 8.38 x 107
    v = 9150 m s-1

    Here I'd like to ask, how come we have equated gravitational potential and kinetic energy? Aren't they two different things?

    Thanks in advance!
     
  2. jcsd
  3. Jul 27, 2010 #2
    OK, I see 3 questions you've asked.

    1) The bold part is the two different distances from the center of the Earth. The first number is the Earth's radius, and the second is the given altitude plus the Earth's radius.

    2) That escape velocity equation is only valid when g is constant. The final height is so far above Earth's surface that gravitational acceleration is no longer -9.8 m/s/s. Which is why you are supposed to find the change in potential energy in part (i) NOT using mg(h1 - h2), but using a more generalized formula.

    3) This is the essence of the law of conservation of energy. If the object begins on Earth's surface with an initial speed and a certain distance from the Earth's center, what type(s) of energy does it have? When the object reaches its maximum altitude, what type(s) of energy does it have? How does its total mechanical energy at launch compare to that at max altitude?
     
  4. Jul 27, 2010 #3
    Thanks for your quick and clear reply ;)


    Nevermind, there still one issue, in the bold part, how come the examiner has taken 6.4 x 10^6 instead of 6.4 x 10^3 ???
     
  5. Jul 27, 2010 #4

    collinsmark

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    Be careful here!

    The gravitational potential (with respect to infinity) of an object of height h above the surface of the earth is

    [tex] \varphi = - G \frac{M}{\left( R_e + h \right)}, [/tex]

    where Re is the approximate radius of the Earth. So if you want to find the gravitational potential difference, solve for φ twice, once when h = 1.3 × 104 km and again when h = 0 km; then find the difference between them.
    But the problem is not asking you to find the escape velocity (which would be ∞ km above the surface of the Earth). It's only asking to to find the initial velocity that gets it up to 1.3 × 104 km above the surface of the earth.
    You can use the escape velocity equation if you are trying to find the velocity required to launch something out to a height of ∞.
    Gravitational potential can be thought of (in certain situations) as potential energy per unit mass. Conservation of energy is where where the kinetic energy fits in.

    The convention is that an object has 0 potential energy when at a distance of ∞. So if closer in, the potential energy is negative. The equation is particularly useful to find differences in potential energy (in which case it matters not what distance you consider 0 energy, because you are only finding the difference).

    Whatever the case, the marking scheme does not equate gravitational potential with kinetic energy. What it's equating is (mass X change in gravitational potential) with change in kinetic energy. Which is the same thing as saying it is equating change in potential energy with change in kinetic energy. Conservation of energy (where air resistance is ignored).

    There is another way to approach this problem that might be more intuitive. Ask yourself, "If I were to drop an object from rest, at a height of 1.3 × 104 km above the surface of the Earth, how fast would it be traveling when it crashed on the surface of the Earth?" That's the same velocity needed to launch it up to the same height. :wink:
     
  6. Jul 27, 2010 #5

    collinsmark

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    Converting km to m?

    [Edit: Oh, HZXAHNLfzjSr. By the way, welcome to Physics Forums!]
     
    Last edited: Jul 27, 2010
  7. Jul 27, 2010 #6
    Oh my god I am so clumsy :D

    Thanks a lot for your detailed reply ;)

    I've got most of it!
     
  8. Aug 9, 2010 #7
    Thanks for the welcome.

    I've returned as I'm still unsure regarding

    ½mv2 = mdeltaφ

    I've thought about it and I just don't understand this equation, especially the "m/mass" part.

    I'd appreciate it if you could explain it to me again :P

    Thanks!
     
  9. Aug 12, 2010 #8

    collinsmark

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    Sorry for the delay, I just returned from travel.

    The

    [tex] \frac{1}{2}mv^2 = m \Delta \varphi [/tex]

    equation means the kinetic energy is equal to the change in potential energy.

    The left hand side, [itex] 1/2 \ m v^2 [/itex], is the expression for kinetic energy.

    The [itex] \Delta \varphi [/itex] is the change in gravitational potential. Multiply that times the mass, and you have the change in gravitational potential energy.

    Equating the two is just an expression of conservation of energy.
     
    Last edited: Aug 12, 2010
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