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Graphing force and acceleration

  1. Aug 19, 2005 #1
    Hey, I'm new to these forums (duh! :biggrin:) and need some help; we have a physics portfolio containing 10 questions of varying difficulties. I'm having some troubles with some of the questions and wouldn't mind some help to guide me in the correct direction! So I'd be grateful if anyone aids me with this.

    The question is:
    7. Draw graphs to show relationship between
    a) FORCE and ACCELERATION for an object in circular motion.
    b) MASS and DISTANCE for forces due to gravitation.
    c) The PERIOD of orbit of a satellite and its ALTITUDE.
    Indicate correct units and symbols of the quantities when labelling the axes for each.

    Well, for part a, I think the graph would have acceleration at the horizontal axes and Force on the vertical axis. The line would probably be straight, but i'm not sure if an object in circular motion changes acceleration at different parts.

    Part b, I think it'd be a straight line because an object wouldn't move unless acted upon by another object. Mass on the vertical axis, distance on the horizontal axis.

    Part c, as the satellite gets closer to eath, it becomes faster. Therefore it spends less time at the lower altitudes. So it'd be like a y=x^2 graph with period on the vertical axes and altitude on the horizontal axis.

    So yup, help me out!

    thanks ! :smile:
  2. jcsd
  3. Aug 19, 2005 #2
    An object in circular motion has a centripetal force vector that is constant in magnitude but changes direction. The acceleration vector is the force vector divided by mass. I wouldn't worry about the direction change.

    This ones sort of vague, but the distance does not affect the mass, so you are right.

    You have the right idea about the speed of orbit with relation to distance. The easiest way to answer this question is to look at the equation for orbit period in relation to distance, and graph the dependance. It is not parabolic though.
  4. Aug 19, 2005 #3
    cool !

  5. Aug 20, 2005 #4
    okay, I have another question:

    10. A science group put in a satellite of mass m kg into a circular earth orbit of radius r. The orbital velocity it needs to remain in this orbit is v. They now put another satellite into a similar orbit at the same altitude. Its mass is 3 times m. What orbital velocity would it need to be given? Give reasons why using Mathematical reasoning.

    I know that it involves this formula, v= 2*pi*r/T and r^3/T^2 = Gm/4*pi^2

    But i'm not sure how to use them.

    Anyone can help? thanks !
  6. Aug 20, 2005 #5
    In a circular orbit, the centripetal force is provided by the gravitational force of the planet. If you equate these two you'll notice something about the mass. Try it out.
  7. Aug 20, 2005 #6
    v = 9.8

    Therefore v^2/r = G*m/r^2
    9.8^2/r = 9.8mr^2

    Therefore if the other satellite is 3 times m
    then 3m = 9.8*3 / r

    ummm, does that even make sense ? :confused:
  8. Aug 20, 2005 #7
    The orbital velocity is: 29.4

    (9.8 * 3)

    I think I'm right, but i'm not 100% sure that v = 9.8
  9. Aug 20, 2005 #8
    Your very first assumption is incorrect. There is no reason to claim v = 9.8. V is the variable you are trying to find. This is what I intended for your first step

    [tex] \frac{m_{obj}v^2}{r} = \frac{GM_{earth}m_{obj}}{r^2} [/tex]
  10. Aug 20, 2005 #9
    oh, okays... my bad :blushing:

    but I'm pretty sure i got it now!

    From the equation:

    mv^2/6.378*10^6 = 9.8*5.974*10^24*m/(6.378*10^6)^2*m
    (cancel the m(object))

    v = 3,029,726,249

    And if the mass was three times as much, it wouldn't matter!

    Thanks! (ummm, warn me if i'm wrong ) :smile:
    Last edited: Aug 20, 2005
  11. Aug 20, 2005 #10
    Well you got the right conclusion, but that velocity is over the speed of light :)

    [tex] \frac{m_{obj}v^2}{r} = \frac{GM_{earth}m_{obj}}{r^2} [/tex]

    [tex] v^2 = \frac{GM_{earth}}{r} [/tex]

    [tex] v = \sqrt{\frac{GM}{r} [/tex]

    There is no mention of the objects mass. Its irrelevant.
  12. Aug 20, 2005 #11
    Oops, I made a mistake...

    I thought G was 9.8, but it's actually the "gravitational constant" which is 6.668*10^-11

    Edit: hehe that's why my V was over the speed of light !!! :rofl:
    Last edited: Aug 20, 2005
  13. Aug 20, 2005 #12
    [tex] 7.903 \times 10^3\ m/s \ is\ the\ orbital\ velocity\ that\ needs\ to\ be\ given. [/tex]

  14. Aug 20, 2005 #13
    I have a question, can you calculate the gravitational field with the formula:

    g = GM/d^2


    I'm getting confused because one website included a negative sign and the other didn't!
  15. Aug 20, 2005 #14


    User Avatar
    Staff Emeritus
    Science Advisor

    The "negative sign" is a matter of convention. Strictly speaking, force is a vector and you have to include the direction as well as the magnitude. If you are talking about gravity pulling things "down" then it is standard to consider the force "down" and set up your coordinate system so that is negative. If you are just talking about the "magnitude" of the force, without the direction, you can ignore the sign.

    By the way, the "gravitational field" refers to the force and its magnitude would be [itex]\frac{GMm}{r^2}[/itex]. Of course, because mg= F, where g is the acceleration due to the gravitational force, you have [itex]g= \frac{GM}{r^2}[/itex]. (Notice that there is no negative sign because I am explicitely talking about the magnitude, not the direction.)
    Last edited: Aug 20, 2005
  16. Aug 22, 2005 #15
    Great! thanks!
  17. Aug 22, 2005 #16
    Okays, this is another question... I believe I answered correctly

    Critically assess the following statement: `Astronauts in a space capsule orbiting the earth at a height of 900 km experience weightlessness: hence the gravitational field due to the earth must be zero at that altitude. Include calculations in your answer.

    Using the formula g = GM/r^2

    We get the answer: [tex] g = 7.52 \ m/s^2 [/tex]

    The astronauts are experiencing weightlessness because they are not used to a gravitational field below 9.8, which is the gravitational field on earth’s surface.

    Does that make sense?
  18. Aug 22, 2005 #17


    User Avatar
    Staff Emeritus
    Science Advisor

    The acceleration due to gravity on earth is about 9.81 m/s2. You calculated that, at a height of 900 km, it is 7.52 m/s2. That's 76% of what they would feel on earth. They are not "experiencing weightlessness" because that's lower than normal. You "feel" your weight when you are standing on the earth, or a floor, or something else that is pushing up on you with the same force as you are pushing down- your weight. When the astronaut is in orbit, the space shuttle is "falling" just as he is. What is pushing up on him/her?

    By the way, it is better to start a new thread for a new question rather than just appending new questions to an old thread. Many people stop looking at threads when they see the orginal question has been answered.
    Last edited: Aug 22, 2005
  19. Aug 22, 2005 #18
    oh okay, thanks for the tip! (I created a new thread for something else

    anyways, the chair he's sitting on is pushing him up?

    actually, i don't think anything really is pushing him up since the space shuttle is, as you said, falling as he is.

    Therefore, he is experiencing weightlessness since nothing is pushing up on him with the same force as he is pushing down.
    Makes sense ?
  20. Aug 22, 2005 #19
    Just because he feels a net force of zero, doesnt mean that there arent forces acting on him..
  21. Aug 23, 2005 #20
    The whole negative sign thing is a stupid thing to get confused about in physics. For each problem, just set up your own coordinate system that makes it easier to do the math, draw out your diagrams with all vectors pointing the right way, and check the vector's direction with your made up coordinate system to determine the sign. If you start doing this now, it'll save you a lot of confusion later.
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