Rearranging gravitational and centripetal equations to derive formulas

In summary, the conversation discusses the relationship between Newton's Second Law and the formula for centripetal acceleration, showing how they can be used to calculate the force and velocity of an orbiting object. It also clarifies that the centripetal force is not a separate force, but rather the gravitational force in this specific scenario.
  • #1
AN630078
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Homework Statement
Hello, I have been tasked with rearranging various equations to derive certain formulas but I do not know whether I have done so correctly, or necessarily using the correct quantities. Would anyone be able to look over my workings and possibly offer some advice?

1. Using Newton’s second law of motion and a formula for centripetal acceleration, write down a formula for the centripetal force on a planet of mass m moving in a circular orbit of radius r with a speed v.

2. Equate your last answer to the force as shown by Newton’s law of universal gravitation, between a planet of mass m at a distance r from the Sun of mass M. Hence show that as r decreases v increases.

3.By using the other formulae for centripetal acceleration, show that T squared is proportional to r cubed.
Relevant Equations
F=ma
ω=2π/T
a=rω^2 = v^2/r
F=GMm/r^2
1. Newton's Second Law states F=ma and the formula for centripetal acceleration is v^2/r
Therefore, F= mv^2/r

Would this be complete, I just feel that I should need to do something further but I am not sure what?

2.F=mv^2/r
Gravitational force = GMm/r^2
Gravity is the cause of centripetal force so these are equal;
mv^2/r=GMm/r^2
Cancel m; v^2/r=GM/r^2
Multiply by r; v^2=GM/r

v^2∝1/r
or v∝1/√r

GM is constant so as v increases r decreases.

3. I would assume that the question here means a= rω^2
Therefore, F=mrω^2
The angular velocity is ω=2π/T

The gravitational force is given by F=GMm/r^2 and since the planets moves in a circular orbit, consider the centripetal force also, F=mω^2r. As both gravitational and centripetal forces act in the same direction, one can equate them to find GMm/r^2=mω^2r.
Note that 'm' cancels; GM/r^2=ω^2r
Divide by r: GM/r^3=ω^2

Substitute the value of the angular velocity into the equation; GM/r^3=(2π/T)^2 which becomes GM/r^3=4π^2/T^2
Multiply both sides by T^2; T^2 GM/r^3=4π^2
Multiply both sides by r^3; T^2 Gm=4π^2*r^3
Divide both sides by GM; T^2 =4π^2*r^3/GM

Since 4π^2/GM is constant T^2 is indeed proportional to r^3.

Thank you very much to anyone who replies 👍
 
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  • #2
AN630078 said:
As both gravitational and centripetal forces act in the same direction, one can equate them
It takes a bit more than that... the gravitational force is supplying the centripetal force, as no other force acts in that direction.

Other than that, all good.
 
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  • #3
haruspex said:
It takes a bit more than that... the gravitational force is supplying the centripetal force, as no other force acts in that direction.

Other than that, all good.
Thank you very much for your reply, do you know I was unsure about that statement before posting so thank you for picking up on that. Do you think I have used the correct equations in the questions specified, e.g. "the other formulae for centripetal acceleration" etc. ? Are there any improvements I could make?
 
  • #4
I'd like to add a few words of clarification too.

Are there any improvements I could make?

Good answer overall but it is incorrect to say:
“both gravitational and centripetal forces act in the same direction"
because it suggests there are 2 different forces acting. There is only one. The centripetal force *IS* the gravitational force in this question.

It is a common error to think that centripetal force is a special type of force, but it is just the name we give to whatever force happens to be acting perpendicular to the velocity. Such a force (be it gravitational, electrical, magnetic or the resultant of several other forces) causes the direction of velocity ot change and the circular path is the result - with the 'centripetal' force pointing towards the centre of the circle.
 
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Related to Rearranging gravitational and centripetal equations to derive formulas

1. What is the purpose of rearranging gravitational and centripetal equations?

The purpose of rearranging these equations is to solve for different variables and understand the relationship between them. By rearranging the equations, we can derive new formulas that can be used to calculate various aspects of gravitational and centripetal forces.

2. How do you rearrange equations to derive new formulas?

To rearrange equations, we use algebraic manipulation to isolate the variable we want to solve for. This involves performing the same operation on both sides of the equation, such as adding, subtracting, multiplying, or dividing, to simplify and solve for the desired variable.

3. Can the rearranged formulas be used for any type of gravitational or centripetal force?

Yes, the rearranged formulas can be used for any type of gravitational or centripetal force as long as the initial equations are applicable to the situation. However, some variables may need to be substituted with known values or measured quantities to accurately use the derived formula.

4. What are some common variables in gravitational and centripetal equations?

Some common variables in these equations include mass, velocity, radius, gravitational constant, and centripetal force. These variables can be rearranged and combined in different ways to derive new formulas for specific situations.

5. How can rearranging equations help in understanding the principles of gravity and circular motion?

Rearranging equations can help in understanding the principles of gravity and circular motion by providing a deeper understanding of the relationships between different variables. By manipulating the equations, we can see how changing one variable affects the others and gain insight into the behaviors of these forces.

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