Earth's varying centripetal acceleration

[Coldie]

Hi,

I've got a problem that I'm working on that is stated as follows:

Earth's elliptical orbit around the sun brings it closest to the sun in January of each year and farthest in July. During what month would Earth's (a) speed-and (b) centripetal acceleration be greatest?

Now, using Kepler's second law, I stated that a greater arc length must be made when the Earth is closer to the sun, and so the speed there must be greater. I was about to put down that acceleration must also increase since $$a_{c} = v^2{}/r$$, but then I realized that the radius has also decreased, and so I wasn't sure how I could determine where on the Earth's elliptical rotation around the sun the centripetal acceleration is greatest. Can someone help me out?

 

[Moose352]

If the radius has decreased, wouldn't that mean that the acceleration has increased since r is on the bottom (of the formula)?

 

[Coldie]

But v has also increased, as was deduced by looking at Kepler's second law...

 

[Coldie]

Come on, someone must be able to explain how to find centripetal acceleration in an elliptical orbit to me. I'd imagine this is a fairly widespread topic considering every planet in the solar system revolves the sun in an elliptical orbit to varying degrees.

 

[Ivan Seeking]

I'm not sure what level you're working at but I think you have what you're looking for. If v is a maximum when r is a minimum, what has happened to a_c?

 

[Coldie]

There's no telling, since the magnitude of both r and v is unknown. If r is sufficiently small, the acceleration will have increased, and if it is sufficiently large, it will have decreased. Or am I missing something?

 

[Ivan Seeking]

Are you thinking that a_c must be constant? We only need the equal areas in equal time, which led you to conclude that v is largest when r is smallest, right? . At this point we know v is large where r is small, so what must be true for a?

What level are you at in school? It's hard to provide the proper level of help without knowing.

 

[Coldie]

Grade 12 Physics right now. It's almost 1AM and I can't believe it's taken me so long to grasp that when you divide by a small number, it gets bigger. God I'm brain-dead sometimes. Thanks very much for the help.

However, I do have another problem that's probably worthy of this forum.

At what distance from Earth, on a line between the centre of Earth and the centre of the moon, will a spacecraft experience zero net gravitational force?

Now, I start out by stating $$r_e{} + r_m{} = 3.84 \times 10^8{}$$, 3.84 x 10^8 being the distance between the Earth and the moon.
I then solve for re: $$r_e{} = 3.84 \times 10^8{} - r_m{}$$

Now I use the gravity equation, $$g = \frac{GM}{r^2{}}$$

Since g must be equal from both planets, I equate the two: $$\frac{GM_m{}}{r_m{}^2{}} = \frac{GM_e{}}{(3.84 \times 10^8{} - r_m{})^2{}}$$, where $$M_m{}$$ is the mass of the moon and $$M_e{}$$ is the mass of the Earth.

Over the course of the last hour, I have found myself utterly unable to solve the equation for rm. I'm pretty sure my equation is correct, and I think it's simply due to my poor algebraic skills that I have been unable to isolate rm. Could someone please tell me how I can do this?

The numbers I'm using are 6.67 x 10^-11 for the gravitational constant, 5.98 x 10^24kg for the mass of the Earth, and 7.35 x 10^22kg for the mass of the moon.

 

[Ivan Seeking]

Also, you are implicitly using two different r's here. On one hand you mean r as the distance from the sun, and in the other, when you look at the acceleration, you mean r as in uniform circular motion.

 

[Ivan Seeking]

I posted late, hang on...

 

[Coldie]

You're right, that equation wouldn't work for the problem I'm given, and I'm completely restricted to using Kepler's laws. I see where I went wrong, and I'm pretty sure I've got the right answer now. Thanks for your help on that one.

[edit]
dontcha hate when that happens?
[/edit]

 

[Ivan Seeking]

Since g must be equal from both planets, I equate the two: $$\frac{GM_m{}}{r_m{}^2{}} = \frac{GM_e{}}{(3.84 \times 10^8{} - r_m{})^2{}}$$, where $$M_m{}$$ is the mass of the moon and $$M_e{}$$ is the mass of the Earth.

Multiply both sides of the equation by the denominators and solve the quadratic equation for r_m

 

[Ivan Seeking]

God I'm brain-dead sometimes.

You will do well in physics!

 

[Coldie]

lol, thanks Ivan.

Attempting to solve... please hold...

[edit]
gah, this may take awhile. If you want to leave, could you leave the answer in spoiler tags or something for me?
[/edit]

 

[Ivan Seeking]

Are you comfortable solving quadratic equations?

 

[Coldie]

Ok, I'm stuck again. I apologize for my ineptitude.

Multiplying both sides by their respective denominators yields:
$$GM_m{}(3.84 \times 10^8{} - r_m{})^2{} = GM_e{}r_m{}^2{}$$

Getting rid of G and expanding $$(3.84 \times 10^8{} - r_m{})^2$$, I get:

$$M_m{}(3.84 \times 10^8{} - r_m{})^2 - M_m{}(3.84 \times 10^8{})r_m{} + r_m{}^2{} - M_e{}r_m{} = 0$$

Plugging in all the largish numbers, I get values that I try to plug into the quadratic formula, but the answer I get that isn't negative has an exponent of 44. I'll try it again, but if you can see something wrong with what I'm doing, please tell me.

[edit]I got 40277514, which sounds feasible. Checking it now...[/edit]

[edit2]No, doesn't work. Making third attempt...[/edit2]

 

[Ivan Seeking]

I'll be on for a little while yet. No hurry. Separate the terms according to the powers of r and reduce to the standard quadratic form.


edit: wait, again I posted just as you did.

 

[Ivan Seeking]

okay, you need to multiply out the (3.84E8 - r_m)² and get r², r, and the constant alone. Remember the form of the quadratic equation: aX² + bX + c = 0

 

[Coldie]

Alright, I'll post exactly what I've got on my sheet right now:
$$-5.98 \times 10^{24}r_m{}^2{} + r_m{}^2{} - 7.35 \times 10^{22}(3.84 \times 10^8{})r_m{} + 7.35 \times 10^{22}(3.84 \times 10^8{})^2{} = 0$$

I calculate all the numbers, and it doesn't seem to work.

Sorry that took so long, LaTeX is cool, but takes me awhile to type out.

[edit]
Wow. This is starting to freak me out

Looking at my second equation up there, I get the impression that it's utterly borked. I did actually do what you're saying, but the way I've got it entered is totally messed. I'll try to redo it, one sec.
[/edit]

[edit2]
$$M_m{}(3.84 \times 10^8{})^2 - M_m{}(3.84 \times 10^8{})r_m{} + r_m{}^2{} - M_e{}r_m{}^2{} = 0$$

All I did was forget to take the rm out of the constant part. AND forget to square the last rm.
[/edit2]

[edit3]
Must go sleep, bye-bye!
[/edit3]

 

[Ivan Seeking]

$$M_m{}(3.84 \times 10^8{})^2 - 2M_m{}(3.84 \times 10^8{})r_m{} + (M_m- M_e{})r_m{}^2{} = 0$$

 

[Coldie]

This should be a lesson to all to never do homework in the AM hours. $$(a + b)^2{} = a^2{} + 2ab + b^2{}$$. I need to be sent back to grade 8. Darn 2.

Is the answer 38113118m?

 

[Ivan Seeking]

plug in your answer and check, slacker.

 

[Coldie]

I totally did! For the gravitational force of the moon I'm getting .003374926m/s^2, and when I plug in 3.84E8 - 38113118 for the radius for the Earth, I get the acceleration as .003333948m/s^2. I've stored all my numbers in my calculator's memory, so I don't understand why there's a discrepancy, and was wondering if the numbers are so large that the calculator's trimming has actually effected the accuracy of the result?

P.S. Are you a teacher?;)

 

[Ivan Seeking]

How many significant figures do you have? I think you may also have a round off error in the mass used for the earth. I would need to google...

 

[Ivan Seeking]

Nope, not a teacher.

 

[Ivan Seeking]

If you have any more trouble I'll check back in the AM, but I'm really drooping now...

 

[Coldie]

Well, thanks for all the help, I think I'll do the same, and I REALLY mean it this time;)

It's probably just my calculator's trimming off the numbers that it gets, anyways. Danke!