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Is the Force of Gravity act as a constant when solving for centripetal acceleration?

  1. Dec 1, 2011 #1
    Hi,

    I know that centripetal force for planetary motion is the same as the force of gravity between that satellite and planet. For example (I know these numbers may be completely unrealistic but just for the sake of easy calculation...) if the mass of the Earth is 1x10^30 kg and the mass of a satellite is 1000kg and the mean distance between their centers is 3000m, then the force of gravity between the two will be about 7.41x10^15 N according to Newton's law of universal gravitation.
    This is also the centripetal force for the satellite in relative circular motion. So 7.41x10^15 N = m(of satellite)v^2/r. Now, if the radius of the orbit of the satellite increases, the force of gravity decreases, the radius in mv^2/r increases and so therefore the velocity decreases.

    But I am wondering how you can explain why my reasoning is wrong if I look at it from this viewpoint. I said that if the radius of orbit increases in mv^2/r, then could we not keep the gravitational force the same by also increasing the velocity? Just like all points on the hand of a clock experience different velocities(higher as you go farther out from the center) but the same centripetal force. But my problem is that I know that the force of gravity MUST decrease with an increasing distance, so is the force of gravity something of a constant? You must solve for it first and then solve for the centripetal acceleration (velocity and radius)?

    Thank you!
     
  2. jcsd
  3. Dec 1, 2011 #2

    D H

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    Re: Is the Force of Gravity act as a constant when solving for centripetal accelerati

    You need to use Newton's universal law of gravitation:

    [tex]F=GMm/r^2[/tex]
     
  4. Dec 1, 2011 #3
    Re: Is the Force of Gravity act as a constant when solving for centripetal accelerati

    I know to use that for the force of gravity but is that what you must solve for first? In the formula, if distance increases, the force decreases, but according to force of gravity equals centripetal force, I could simply increase the velocity and not change the force of gravity (which I am having trouble understanding because I know that the force of gravity must decrease)
     
  5. Dec 1, 2011 #4
    Re: Is the Force of Gravity act as a constant when solving for centripetal accelerati

    Gravity is purely a function of Mass and Distance based of its universal form. The centripetal force you speak of f = mv^2/r is a fictitious force. Meaning it is not true force. That formula was derived such that newtons laws would work with circular motion. Probably doesn't answer your question completely but it's a step in the right direction...

    Further, Centripetal force and gravity are both in the same direction. If I were to accelerate a Baseball downward at 9.81m/s which happens to be the same as the acceleration due to gravity you would not say that the two forces were the same simply that the two forces shared the same magnitude. So, though the two force may be equal in magnitude and direction that doesn't mean they are the same force. So if you increase r, gravity will drop off at a rate of 1/r^2 and cent Force will drop off at a rate of 1/r. If you want the two forces to remain with the same, then then you would need to decrease the velocity in order for the maintain the same relation of the magnitudes. However, this does not imply that they are the same force, they have the same magnitude.
     
    Last edited: Dec 1, 2011
  6. Dec 1, 2011 #5
    Re: Is the Force of Gravity act as a constant when solving for centripetal accelerati

    I am sure that centripetal force is not a fictitious force. We have learned about that for a week or two in physics and I have seen many websites derive it. I think you may be misusing the term. CentriFUGAL force is a fictitious force not centripetal force.

    And I am also 100% sure that the force of gravity provides the centripetal force for planetary motion around a central mass. If you ask you teacher, look it up, or even think about it in a mean circular motion sense, the force of gravity provides the centripetal force.
     
  7. Dec 1, 2011 #6
    Re: Is the Force of Gravity act as a constant when solving for centripetal accelerati

    Please explain where F=GMm/r^2 provides a rotational force. I see one dimensional movement in the r direction no theta involved. Irregardless of the fictitious force issue, my example still stands.
     
  8. Dec 1, 2011 #7
    Re: Is the Force of Gravity act as a constant when solving for centripetal accelerati

    Centripetal force is the name given to any force which curves a particle's path, the equation given is the magnitude of the force required (at a right angle to the particle's velocity) to keep the particle in circular motion. I really don't know how you plan to explain orbits if you insist that gravity cannot provide a 'rotational force', what other force do you attribute the curved paths of satellites to?

    The lack of a [itex]\theta[/itex] in the formula for gravity doesn't restrict it from curving a path, it only implies that angular momentum is conserved in that system relative to the object being orbited.
     
  9. Dec 1, 2011 #8
    Re: Is the Force of Gravity act as a constant when solving for centripetal accelerati

    I agree with JHamm...the object or satellite will always want to travel in the direction and magnitude in which it is already traveling according to Newton's first law. The satellite orbiting a planet had an initial velocity and is trying to move forward. No force is acting horizontally on this motion so it will continue (tangent to the circle). But gravity is pulling on the object perpendicular to it horizontal movement, causing a curved path. Here is a good derivation of centripetal force which equals the gravitational force for planetary orbits:

    http://dev.physicslab.org/Document....me=CircularMotion_CentripetalAcceleration.xml

    remeber that net force is mass times acceleration. In centripetal force the mass times acceleration (change in velocity over time or v^/r for a circle) continues to fit Newton's second law.
     
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