Why do the centripetal and gravitational force equal each other in orbit??Also...


by MakeItThrough
Tags: circular, friction, gravitation, motion, uniform
MakeItThrough
MakeItThrough is offline
#1
Oct19-11, 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
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Mindscrape
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#2
Oct19-11, 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.
PeterO
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#3
Oct19-11, 08:46 PM
HW Helper
P: 2,316
Quote Quote by MakeItThrough View Post
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
Simplest acceleration is this:

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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