Earth's varying centripetal acceleration

In summary: I said Mm(3.84 x 108 - rm)^2, when it should be Mm(3.84 x 108 + rm)^2.Now I can solve it, I think.In summary, a student is seeking help on determining when Earth's speed and centripetal acceleration would be greatest in its elliptical orbit around the sun. They use Kepler's second law and realize that acceleration must also increase when the radius decreases, but they are unsure how to determine the specific point in the orbit where the acceleration is greatest. They also ask for help on a problem involving a spacecraft's distance from Earth and the moon where it experiences zero net gravitational force. The student
  • #1
Coldie
84
0
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 [tex]a_{c} = v^2{}/r[/tex], 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?
 
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  • #2
If the radius has decreased, wouldn't that mean that the acceleration has increased since r is on the bottom (of the formula)?
 
  • #3
But v has also increased, as was deduced by looking at Kepler's second law...
 
  • #4
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
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
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
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
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 [tex]r_e{} + r_m{} = 3.84 \times 10^8{}[/tex], 3.84 x 10^8 being the distance between the Earth and the moon.
I then solve for re: [tex]r_e{} = 3.84 \times 10^8{} - r_m{}[/tex]

Now I use the gravity equation, [tex]g = \frac{GM}{r^2{}}[/tex]

Since g must be equal from both planets, I equate the two: [tex]\frac{GM_m{}}{r_m{}^2{}} = \frac{GM_e{}}{(3.84 \times 10^8{} - r_m{})^2{}}[/tex], where [tex]M_m{}[/tex] is the mass of the moon and [tex]M_e{}[/tex] 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
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
I posted late, hang on...
 
  • #11
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]
 
  • #12
Since g must be equal from both planets, I equate the two: [tex]\frac{GM_m{}}{r_m{}^2{}} = \frac{GM_e{}}{(3.84 \times 10^8{} - r_m{})^2{}}[/tex], where [tex]M_m{}[/tex] is the mass of the moon and [tex]M_e{}[/tex] is the mass of the Earth.

Multiply both sides of the equation by the denominators and solve the quadratic equation for rm
 
  • #13
Coldie said:
God I'm brain-dead sometimes.

You will do well in physics! :biggrin:
 
  • #14
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]
 
Last edited:
  • #15
Are you comfortable solving quadratic equations?
 
  • #16
Ok, I'm stuck again. I apologize for my ineptitude.

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

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

[tex]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[/tex]

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]
 
Last edited:
  • #17
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.
 
  • #18
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
Alright, I'll post exactly what I've got on my sheet right now:
[tex]-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[/tex]

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 :-p

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]
[tex]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[/tex]

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:
  • #20
[tex]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[/tex] :smile:
 
  • #21
This should be a lesson to all to never do homework in the AM hours. [tex](a + b)^2{} = a^2{} + 2ab + b^2{}[/tex]. I need to be sent back to grade 8. Darn 2.

Is the answer 38113118m?
 
  • #22
plug in your answer and check, slacker. :biggrin:
 
  • #23
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?;)
 
  • #24
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...
 
  • #25
Nope, not a teacher.
 
  • #26
If you have any more trouble I'll check back in the AM, but I'm really drooping now...
 
  • #27
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!
 

Related to Earth's varying centripetal acceleration

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration that an object experiences as it moves in a circular path. It is always directed towards the center of the circle and its magnitude depends on the object's speed and the radius of its circular path.

2. How does Earth's varying centripetal acceleration affect us?

Earth's varying centripetal acceleration is the reason why we experience different weight at different locations on Earth. This is due to the fact that the Earth's rotation causes a change in our distance from the center of the Earth, which in turn affects the acceleration we experience.

3. What causes Earth's varying centripetal acceleration?

Earth's varying centripetal acceleration is primarily caused by the Earth's rotation around its axis and its orbit around the sun. These movements result in a change in our distance from the center of the Earth, causing a change in the centripetal acceleration we experience.

4. How does Earth's varying centripetal acceleration affect gravity?

Earth's varying centripetal acceleration does not directly affect gravity, as gravity is a force that is always directed towards the center of the Earth. However, it does play a role in determining our weight, as the variation in our distance from the center of the Earth affects the gravitational force we experience.

5. Can Earth's varying centripetal acceleration be measured?

Yes, Earth's varying centripetal acceleration can be measured using various techniques and instruments. For example, measuring the duration of a pendulum swing at different latitudes can provide information about the difference in centripetal acceleration due to the Earth's rotation.

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