Variation of gravitational force with calculus

In summary, the conversation discusses Newton's universal law of gravitation and how to compute the instantaneous acceleration and orbital velocity using calculus. The participants also consider the conditions of circular orbits and the approximation of one object remaining stationary. Ultimately, the acceleration can be found by dividing the magnitude of the force by the mass of the object, and the velocity can be calculated by setting the centripetal acceleration equal to the gravitational acceleration and solving for velocity.
  • #1
albertrichardf
165
11
Hi all,

This is Newton's universal law of gravitation:

F = GMm/r2, where r is the distance between the centre of the two bodies.

Therefore, considering two objects in mutual gravitational acceleration, with only linear motion and acceleration, they shall be moving in closer and closer. Since the force is inversely proportional to the distance between the two bodies centre, it will increase as r decreases. Therefore as they move gravitational force gets stronger and stronger.

The question is, how to compute the instantaneous acceleration in those conditions using calculus?

Also, considering the two objects again, how would I compute the velocity needed for mass m to enter in orbit around mass M?

Thanks for any answers
 
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  • #2
The instantaneous acceleration is pretty easy to find simply by applying Newton's second law. What's Newton's second law tell you? This should not require any calculus.

Computing the orbital velocity requires knowledge of which orbit you are talking about. Different orbits require different velocities.

In the case of elliptical orbits, there is no 1 velocity that is the "orbit velocity" because by Kepler's 2nd law, the orbiting object moves fastest when it is closest to the gravitating body.

Circular orbits have 1 velocity, and they depend on the distance R from which you are orbiting. The problem is easier still when one object (say M) is much more massive than the other object so we can make an approximation that M just stays stationary (just an approximation).

Which conditions are you considering?
 
  • #3
So my acceleration at any point would be the magnitude of the force at the distance r divided by the mass of object 2?
The conditions I am considering are a circular orbit of distance r, where mass M around which mass m orbits can be approximated to remain stationary.

Thanks.
 
  • #4
I believe in that case you simply find the magnitude of the force and calculate the acceleration. Since you have a circular orbit, the distance between the two objects will not change and you will have no variation in either the force or the acceleration.
 
  • #5
Albertrichardf said:
So my acceleration at any point would be the magnitude of the force at the distance r divided by the mass of object 2?
The conditions I am considering are a circular orbit of distance r, where mass M around which mass m orbits can be approximated to remain stationary.

Thanks.

You should be careful to specify which object you are talking about. The acceleration of object 2 will be the force you wrote down divided by the mass of object 2.

For a circular orbit, you can use the above fact AND the fact that for circular motion the centripetal acceleration is v^2/r. Using these two facts, you can set the two accelerations equal to each other and solve for the velocity. This should also only require algebra. Although deriving a=v^2/r takes a little bit of vector manipulation. You should be able to find that in any introductory physics textbook.
 
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FAQ: Variation of gravitational force with calculus

What is the formula for calculating the variation of gravitational force using calculus?

The formula for calculating the variation of gravitational force using calculus is F = G (m1m2/r^2), where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

How does the distance between two objects affect the gravitational force?

The force of gravity between two objects is inversely proportional to the square of the distance between them. This means that as the distance between two objects increases, the force of gravity between them decreases.

What is the relationship between mass and gravitational force?

The force of gravity between two objects is directly proportional to the product of their masses. This means that as the mass of one or both objects increases, the force of gravity between them also increases.

What is the significance of using calculus in understanding gravitational force?

Calculus is essential in understanding the variation of gravitational force because it allows us to calculate the exact force between two objects at any given distance or mass. It also helps us to analyze how this force changes with respect to different variables and make accurate predictions.

Can calculus be used to explain other physical phenomena besides gravitational force?

Yes, calculus is a powerful mathematical tool that can be used to explain various physical phenomena, such as motion, electricity, and heat transfer. It allows scientists to model and analyze complex systems and make predictions about their behavior.

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