Kepler's Third Law vs Conservation of angular momentum

[ShaunPereira]

Homework Statement

If the earth were to triple its present its present distance from the sun then the number of days in one year will be

Relevant Equations

T^2 α r^3
I1ω1= I2ω2

The classic way to go about this problem would be to use Kepler's laws and thus find the new time period of earth.
However I encountered this question in a test on rotational motion which deals with conservation of angular momentum.
The equation used here would be I₁ω₁= I₂ω₂
Replacing I with MR² and ω with 2π/T we get,
MR₁²2π/T₁= MR₂²2π/T₂
which gives us T α R² which is different from what Kepler's third law gives us.

Why shouldn't I use conservation of angular momentum in a problem like this even when it makes sense to do so as there is no net torque acting on the planet?
Is this because the force acting on the planet is gravity which is an inverse square force or is there something else that I am not able to understand?

 

[Drakkith]

Imagine we do the reverse and cut Earth's orbital diameter by a third. Let's give the Earth a kick in the opposite direction of its current orbital velocity such that its perigee now matches the new orbit. The Earth begins to fall towards the Sun, and by the time it reaches perigee we have to slow it again because it's going to fast to stay in a circular orbit. So to get it to its new orbit we had to slow it down twice and remove angular momentum twice.

When increasing the orbital diameter of an object we have to give it a kick at least twice in the direction of its orbital velocity, adding angular momentum.

So changing orbits doesn't conserve angular momentum. In fact, it requires a change in angular momentum.

Why shouldn't I use conservation of angular momentum in a problem like this even when it makes sense to do so as there is no net torque acting on the planet?

It does require a net torque to act on the planet in order to slow or speed it up to change its orbit. You can't magically transport the Earth to its new orbit. As to why the math comes out the way it does, I can't answer that.

 

[ShaunPereira]

Imagine we do the reverse and cut Earth's orbital diameter by a third. Let's give the Earth a kick in the opposite direction of its current orbital velocity such that its perigee now matches the new orbit. The Earth begins to fall towards the Sun, and by the time it reaches perigee we have to slow it again because it's going to fast to stay in a circular orbit. So to get it to its new orbit we had to slow it down twice and remove angular momentum twice.

When increasing the orbital diameter of an object we have to give it a kick at least twice in the direction of its orbital velocity, adding angular momentum.

So changing orbits doesn't conserve angular momentum. In fact, it requires a change in angular momentum.It does require a net torque to act on the planet in order to slow or speed it up to change its orbit. You can't magically transport the Earth to its new orbit. As to why the math comes out the way it does, I can't answer that.

Well what I wanted was an explanation as to why the mathematics(equations) turn out the way they did.
To change a planets orbit we would indeed have to change the angular momentum and as you pointed out it would have to be done twice to achieve the desired orbit.

However I have encountered similar problems like for example,

A body rotating in a circular path of radius r with angular velocity ω. What would the angular velocity be if we were to increase the radius to 2r

In a problem like this angular momentum would be conserved and thus the final angular velocity would be ω/2.

Wouldn't the same argument that we would have to change the angular momentum of the body to achieve the desired orbital radius apply here too or has it got to do with the fact that in a problem
that deals with planets we need to take into account that the force of gravity acts on a planet thus indicating that Kepler's law be used and the law of conservation of momentum be discarded?

​

 

[jbriggs444]

Why shouldn't I use conservation of angular momentum in a problem like this even when it makes sense to do so as there is no net torque acting on the planet?

It matters how you decide to put the Earth into its new orbit.

If you simply wave your magic wand and move it from a 93 million mile orbital radius to a 279 million mile orbit while retaining its original tangential velocity then you've tripled the angular momentum. Unfortunately, the original velocity is too high for a circular orbit at this new radius. You have an elliptical orbit which is not what you want.

If you decide to wave your wand and conserve angular momentum this time, you'll need to reduce tangential velocity by a factor of three. This velocity is too low for a circular orbit at this new radius. Again, you'll have an elliptical orbit which is not what you want.

There are similar problems if you decide to conserve energy.

The most economical way to boost the Earth to triple its orbital radius is with a Hohmann transfer orbit. You strap rockets onto the Earth, thrust forward as hard as you can and boost it into an elliptical orbit with a low point at the original circular orbit and a high point at the target circular orbit. You wait half a [elliptical-]year for the high point in the orbit and you thrust again, as hard as you can in the forward direction to circularize the orbit.

Both burns increase velocity and angular momentum. In spite of this, the new orbital velocity is less than the original orbital velocity. It went way down during the elliptical transfer. Net orbital energy has increased. Angular momentum has increased.

 

[ShaunPereira]

Right, so if I have understood you correctly there is no way to conserve angular momentum and also retain the original shape of the orbit. In order to do so we would first have to accelerate the Earth thus creating an elliptical orbit and then wait till we reach apoapsis and burn prograde until we reach a near circular orbit. Since both times we have increase its velocity and thus kinetic energy, angular momentum has increased and hence not conserved.

However I still have one doubt that is unsolved

However I have encountered similar problems like for example,

A body rotating in a circular path of radius r with angular velocity ω. What would the angular velocity be if we were to increase the radius to 2r

In a problem like this angular momentum would be conserved and thus the final angular velocity would be ω/2.

Wouldn't the same argument that we would have to change the angular momentum of the body to achieve the desired orbital radius apply here too or has it got to do with the fact that in a problem
that deals with planets we need to take into account that the force of gravity acts on a planet thus indicating that Kepler's law be used and the law of conservation of momentum be discarded?

In this problem the body is replaced by a small point mass and the gravitational force is replaced by say tension. Why wouldn't the same argument that applies to Earth no apply to a simple system like this?

 

[jbriggs444]

In this problem the body is replaced by a small point mass and the gravitational force is replaced by say tension.

So now we have a ball on a string running through a hole in the middle of a frictionless, round table. It is moving in a clrcle with radius 1. We slowly let out the string so that the string now has radius 3.

You ask whether angular momentum is conserved.

Yes. This time angular momentum is conserved. The string does not apply an inverse square force. We have contrived for it to provide as much or as little force as is required for a particular path to result. Our desire for a circular orbit presents no requirement that any particular speed result from the maneuver. The string applied no torque (about the hole) to the ball at any time, nor did anything else in the setup.

Energy and tangential velocity are not conserved. While the string was being let out, there was a radial component to the ball's trajectory against which the string acted. Negative work was done on the ball by the string. So the ball lost energy and speed.

 

[ShaunPereira]

The string does not apply an inverse square force

This is exactly what I was trying to imply.

Homework Statement:: If the Earth were to triple its present its present distance from the sun then the number of days in one year will be
Relevant Equations:: T^2 α r^3
I1ω1= I2ω2

Is this because the force acting on the planet is gravity which is an inverse square force

So the force acting on a planet which is gravity, being an inverse square force has a required tangential velocity for a circular orbit to be formed. Anything less or more would lead to an elliptical orbit.

However the force acting on a small ball which here would be tension, as it is not an inverse square force there is no such requirement for velocity to be a certain value in order for the orbit to be circular. It will always be circular no matter the velocity provided the tension in the rope is able to sustain the circular motion

Right?

 

[jbriggs444]

Right?

Right.

 

[ShaunPereira]

Thank you!