Time period of an object traveling in an elliptical orbit

[Saptarshi Sarkar]

Homework Statement

A body of mass 1kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 1000 kg m²/s. What is the time period of the motion?

Relevant Equations

Angular Momentum : mvr = constant
Equation of ellipse : $\frac 1 r = \frac {1 + εCosθ} l$ where r = distance from foci and l = length of latus rectum

I know that the angular momentum of the particle orbiting in an elliptical path is constant and due to which the particle speeds up near the foci when r is small.

But, I cannot figure out how to calculate the time period of rotation. I can do the same for an ellipse by taking mv²/r = central force, but not able to find the equation required for the ellipse.

 

[gneill]

Kepler's second law is the key. It is essentially a statement about angular momentum, and "specific angular momentum" in particular.

What's the total area of your ellipse?

 

[Saptarshi Sarkar]

Kepler's second law is the key. It is essentially a statement about angular momentum, and "specific angular momentum" in particular.

What's the total area of your ellipse?

$πab = π*1000*100 = 100000π$

 

[PeroK]

$πab = π*1000*100 = 100000π$

At this level you shouldn't just answer the direct question and wait for the next hint. You have to think about how knowing the area of the ellipse helps you solve the problem.

In fact, @gneill already gave you a further hint: Kepler's second law.

The posts that homework helpers give are designed to make you think.

 

[Saptarshi Sarkar]

At this level you shouldn't just answer the direct question and wait for the next hint. You have to think about how knowing the area of the ellipse helps you solve the problem.

In fact, @gneill already gave you a further hint: Kepler's second law.

The posts that homework helpers give are designed to make you think.

Sorry about that. I think I found a way to calculate the time period with the help of @gneill and Keplar's second law.

I can divide the total area of the ellipse by the areal velocity $\frac {r×b} 2$ which is equal to $\frac {mr×v} 2 = \frac L {2m}$. So, $T = \frac {2\pi abm } L$

 

[gneill]

Well done!

The specific angular momentum, usually denoted by $h$, and is the angular momentum per unit mass for an orbiting body, has units $m^2/s$. That's area per second (or in general, unit area per unit time).

If you have the angular momentum, $L$, for an orbiting object and you also know it's mass $m$, then the specific angular momentum is $h = \frac{L}{m}$. So what you've done in your calculation is divide the total area of the ellipse by the specific angular momentum, giving you the period of the orbit.

 

[Saptarshi Sarkar]

So what you've done in your calculation is divide the total area of the ellipse by the specific angular momentum, giving you the period of the orbit.

Shouldn't the areal velocity be half the specific angular momentum? Because when I calculated the area swept per unit time of an ellipse it came out to be $\frac {r×v} 2$ (area of triangle made up of vectors r and v)

 

[gneill]

By your own derivation you found:

$T = \frac {2\pi abm } L$

Which is:

$2\pi ab \cdot \frac{m}{L}$ and since $\frac{m}{L} = \frac{1}{h}$ you end up with:

$T = \frac{2 \pi a b}{h}$

 

[Saptarshi Sarkar]

By your own derivation you found:

$T = \frac {2\pi abm } L$

Which is:

$2\pi ab \cdot \frac{m}{L}$ and since $\frac{m}{L} = \frac{1}{h}$ you end up with:

$T = \frac{2 \pi a b}{h}$

Yes, but the $2πab$ is twice the area of the ellipse.

So what you've done in your calculation is divide the total area of the ellipse by the specific angular momentum, giving you the period of the orbit.

This says that $T = \frac {πab} h$

 

[gneill]

You're right. You need to divide the area by h/2.

 

[PeroK]

We have $\frac{dA}{dt} = \frac{L}{2m} = \frac h 2$.

Basically, the angular momentum would be the area of a rectangle $v_{\perp} \times r$, but the area of the ellipse being mapped out is only a triangle.

PS I see we are all in agreement in any case.