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Universal Gravitational constant

  1. Oct 2, 2006 #1
    At the Earth's surface a projectile is launched straight up at a speed of 10.1 km/s. To what height will it rise?
    Universal gravitational constant = 6.673e-11 N m^2/kg^2
    Radius of the earth = 6.370e+6 m
    Mass of the earth = 5.980e+24 kg

    I know to use the equation U= - (Gm1m2)/ r
    but I haven't been getting the right answer
    originally I took into consideration that U=1/2mv^2 and set the two equations equal to each other and solved for r which I took to be R+h.
    R = radius of earth and h being the rest of the distance starting from the earth's surface.

    I would appreciate any help on this it's driving me crazy!! :uhh:
     
  2. jcsd
  3. Oct 2, 2006 #2
    The gravitational constant here is useless, well..we dont really need it. We already know that g=9,8, with that you should be able to find it.
     
  4. Oct 2, 2006 #3

    chroot

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    Absolutely incorrect. The acceleration due to gravity is only 9.8 m/s^2 on the surface of the Earth -- as you go up in altitude, the acceleration due to gravity becomes smaller. An initial velocity of 10 km/s is too fast to permit the approximation that g is everywhere constant.

    - Warren
     
  5. Oct 2, 2006 #4

    chroot

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    To the OP,

    Use the equation for the force felt by the projectile due to the force of gravity. Use that to find its acceleration. Use the acceleration, along with its initial altitude and velocity, to solve its kinematic equation for velocity = 0. That corresponds to its highest altitude: the point at which it stops going up and starts coming down.

    - Warren
     
  6. Oct 2, 2006 #5

    Pythagorean

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    EDIT: This was under the faulty assumption that the speed was 10.1 m/s.

    How to answer your question:

    1) Draw a free body diagram for the projectile (make it a square or a ball or something simple) and draw the forces on it.

    2) Newton's Second law: Once you've drawn your free body diagrams, and labeled the forces on it, you can set up an equation in the x and y directions to represent the directional components of that force.

    3) Setup your equations based on the diagram. There should on be a y component, since the ball is thrown straight up with no wind or air drag.

    4) There's two different things you can do with the equations. You can either use calculus to find your distance and velocity equations, or if you're in an algebra class, look up the equations of kinetics or the equations of motion, and use what you know (like initial velocity and final velocity will have a relationship with acceleration), and since you want your final velocity to be 0 (at the top of it's throw, it comes to a stop for an instant before it switches direction), you can find the time it takes for initial velocity to go to 0 given the the way it's acclerated downward by gravity.
     
    Last edited: Oct 2, 2006
  7. Oct 2, 2006 #6

    Pythagorean

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    oh my, I didn't realize it was going 10 KILOmeters per second.
     
  8. Oct 2, 2006 #7

    chroot

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    Actually, the OP's initial argument was the easiest: it uses the conservation of energy.

    At the surface of the earth, it has kinetic energy 1/2 mv^2. In reaching its maximum altitude, it converts all of this kinetic energy to gravitational potential energy, (-GMm)/r.

    Upon closer inspection, it really looks like your method for determining altitude is totally correct -- yet you say you're getting the wrong answer. What answer did you get, and what leads you to believe it's wrong?

    - Warren
     
  9. Oct 2, 2006 #8
    I set 1/2mv^2=(-GMm)/(R+h)
    1/2m(1.01e4m/s)^2= (-6.673e-11)(5.98e24)m/(6.730e6 + h)
    I solved for h and got 1.45e6m

    This homework question is on a computer program that tells me whether It's right or not. That's why it's so frustrating.
     
  10. Oct 2, 2006 #9

    chroot

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    Looks like you've got the right idea, but your final answer is incorrect. Perhaps you're just making some kind of arithmetic mistake?

    - Warren
     
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