# Centripetal and Gravitational force

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?

• Why is it not true for "small objects"? Apr 15, 2017 at 20:48
• 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. Apr 15, 2017 at 21:07
• 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.... Apr 15, 2017 at 22:28
• I made the question up so there might be errors but if you find one please show me Apr 15, 2017 at 22:30
• 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. Jul 15, 2021 at 19:33

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.