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If F = $mv^2\over r$..... Centripetal force And $F = \frac{Gm_1m_2}{r^2}$....... Gravitational force

Then $\frac{mv^2}{r} = \frac{Gm1m2}{r^2}$ .... But this is not true for all cases especially small objects....

Assuming A = $mv^2\over r$ and B = $Gm1m2\over r^2$, From the above we have that A = B and mathematics states that this should be true for all instance.

If A is 1, then B is surely 1 and if A is infinity, B must be equal to infinity....

Why doesn't Centripetal force equal Gravitational force for small objects?

Please correct me of any mistakes I may have made...

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  • $\begingroup$ Why is it not true for "small objects"? $\endgroup$
    – caverac
    Apr 15, 2017 at 20:48
  • $\begingroup$ Your $A$ must be either $\frac{m1\,v^2}{r}$ or $\frac{m2\,v^2}{r}$ and so when put equal to $\frac{G\,m1\,m2}{r^2}$ one of the masses cancels out. $\endgroup$
    – Farcher
    Apr 15, 2017 at 21:07
  • $\begingroup$ Yes... But its not true for small objects as in this example.... An object has a mass of 2kg and another object which moves around it, a mass of 0.2kg, the distance between them is 1m, the velocity of the second object is 2 ms^-1 when you solve, gravitational force = 2.6x10^-11N but for centripetal force, you have 0.4N.... $\endgroup$
    – Phease
    Apr 15, 2017 at 22:28
  • $\begingroup$ I made the question up so there might be errors but if you find one please show me $\endgroup$
    – Phease
    Apr 15, 2017 at 22:30
  • $\begingroup$ In the question you made up, you added circular motion, so you need to bring in the force say, $F$, that is helping gravity in generating the circular motion. Once you have it, $F + F_{gravity} = \frac{mv^2}{r}$, and there is no discrepancy. $\endgroup$
    – manisar
    Jul 15, 2021 at 19:33

2 Answers 2

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The problem is that the equation you are citing is not a fundamental truth like F=ma, which always holds in Newtonian mechanics. So not all equations are created equal-- some represent inescapable laws (which just means it is much more difficult to escape their realm of applicability), whereas others only hold in quite special situations. Motion in a circle at constant speed is one of those special cases, and only then will A=B in your equation. So the reason it is easy for A to not equal B is that it is easy to violate the requirements of that equation-- it is easy for the objects in question to not be going in a circle, and it is easy for them to not have the only forces on them be gravity.

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I think that by small things you mean "Why don't we see a grape (lower mass) orbiting a watermelon (higher mass)?". It would be really cool to see something like that. And that's the same line of thought Newton developed. He thought that if the earth exerts a force over an apple, this apple exerts the same force on earth proportionally to its mass. Therefore it is true for small objects. We just don't see a grape orbiting a watermelon because they are both on earth, and earth's gravitational force makes these objects' gravitational force negligible.

In your case, you won't find the same value because when you assume this equality $\frac{mv^2}{r} = \frac{Gm_1m}{r^2}$ you have to make sure the problem is not set inside a gravitational field(earth), what would change the gravitational force resultant.

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