Why do the centripetal and gravitational force equal each other in orbit??Also...by MakeItThrough Tags: circular, friction, gravitation, motion, uniform 

#1
Oct1911, 07:32 PM

P: 13

1. The problem statement, all variables and given/known data
Say for example, a problem wants us to find the mass of a planet. It gives us a satellite that orbits that planet with a radius of R and a period T. Now, I know how to solve this problem. You must set Fc = Fg. But what I do not know is why the centripetal and gravitational force of these two objects must equal each other. Also, a similar problem to that is one like this: When you take your 1200 kg car out for a spin, you go around a corner of radius 57.6 m with a speed of 15.2 m/s. The coefficient of static friction between the car and the road is 0.84. Assuming your car doesn't skid, what is the force exerted on it by static friction? Again, I already know how to solve this. You must set Fc = Ff ... mv^2 / r = Ff, and then you just plug in the given values into the mv^2 / r and that is your answer. I do not know why in this case the centripetal force and the static friction must equal each other. If someone could please explain this to me, I would feel so much better while taking the test tomorrow... My teacher goes through this stuff extremely fast. 2. Relevant equations V = 2piR / T Fc = mv^2 / r Fg = Gm1m2 / r^2 



#2
Oct1911, 07:48 PM

P: 1,877

Newtons second law, sum of forces equals mass times acceleration. The force you have is gravity, and the acceleration is centripetal acceleration.




#3
Oct1911, 08:46 PM

HW Helper
P: 2,316

The gravitational force is the ONLY force acting on the Satellite. The Centripetal force is the force we would need to have if the satellite is to travel in a circular path. The satellite DOES travel in a circular path, so the Graitaional force existing happily equals the centripetal force we need. Gravtational force is a real force. Centripetal force is a desired/necessary force. Another example: Why CAN'T you ride a motorbike in a circle of radius 10m at 150 km/h on flat ground? Simple, you could calculate the centripetal force NEEDED for that to happen, but when you add up [as vectors] all the forces acting  gravity, reaction force, friction,... They just don't add up to the necessary force, so the situation just can't happen. In the case of a satellite, the available force [gravity] happens to provide the required force, so its circular motion is maintained. EDIT: Oh, and in the case of the car  it must have been on flat ground also. Most roads have a small degree of banking so that would have contributed, and if the banking was steep enough  like at a velodrome  you mightn't need friction at all. 


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