# Earth's varying centripetal acceleration

1. Dec 11, 2004

### 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?

2. Dec 11, 2004

### Moose352

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

3. Dec 11, 2004

### Coldie

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

4. Dec 12, 2004

### 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.

5. Dec 12, 2004

### Ivan Seeking

Staff Emeritus
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 ac?

6. Dec 12, 2004

### 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?

7. Dec 12, 2004

### Ivan Seeking

Staff Emeritus
Are you thinking that ac 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.

8. Dec 12, 2004

### 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.

9. Dec 12, 2004

### Ivan Seeking

Staff Emeritus
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.

10. Dec 12, 2004

### Ivan Seeking

Staff Emeritus
I posted late, hang on...

11. Dec 12, 2004

### 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.

dontcha hate when that happens?
[/edit]

12. Dec 12, 2004

### Ivan Seeking

Staff Emeritus
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 rm

13. Dec 12, 2004

### Ivan Seeking

Staff Emeritus
You will do well in physics!

14. Dec 12, 2004

### Coldie

lol, thanks Ivan.

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

Last edited: Dec 12, 2004
15. Dec 12, 2004

### Ivan Seeking

Staff Emeritus
Are you comfortable solving quadratic equations?

16. Dec 12, 2004

### 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.

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

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

Last edited: Dec 12, 2004
17. Dec 12, 2004

### Ivan Seeking

Staff Emeritus
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. :yuck:

18. Dec 12, 2004

### Ivan Seeking

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

19. Dec 12, 2004

### 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.

Wow. This is starting to freak me out :tongue2:

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]

Last edited: Dec 12, 2004
20. Dec 12, 2004

### Ivan Seeking

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