How Do You Calculate Sun's Gravitational Acceleration at an Asteroid's Orbit?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the gravitational acceleration of the sun at the distance of an asteroid's orbit, specifically at 1.6 x 1011 m. The context includes the asteroid's orbital period of 398 Earth days, and participants are exploring the application of gravitational concepts and equations related to circular motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the original poster's calculations involving velocity and centripetal force, questioning the assumptions of uniform circular motion. There are suggestions to apply Newton's law of gravitation and to clarify the mass of the sun and gravitational constant. Some participants also point out potential errors in the period of revolution calculation.

Discussion Status

The discussion is active, with participants providing guidance on necessary corrections and considerations. There is an emphasis on clarifying assumptions regarding the asteroid's motion and the appropriate application of gravitational laws, but no consensus has been reached on the correct approach.

Contextual Notes

Participants note the importance of stating assumptions about the asteroid's motion, particularly regarding uniform circular motion, and highlight the need for accurate values such as the mass of the sun and the gravitational constant.

pumpernickel
Messages
8
Reaction score
0
Hi. I am a little stuck and I would appreciate some help.

What is the acceleration due to gravity of the sun at the distance of 1.6 X 10^11 m? The asteroid revolves around the sun in 398 Earth days.

2. Homework Equations :

F= (m*V^2)/r3. The Attempt at a Solution :

First I found the circumference: 1.00 * 10 ^12 m
Converted 398 to seconds --> 85968000 sec

So I used that to come up with a velocity --> 1.00 * 10 ^12 m/85968000 sec = 11694 m/s

And did F= (V^2)/r and came up with .000855

I know I am doing this wrong but I just don't know what to do.

Thanks for your help!
 
Physics news on Phys.org
Not sure why you think you are doing this wrong except maybe you assumed that the asteroid is moving in uniform circular motion which it probably is not in reality. Even though it might be a good approximation the exact answer is Newton's law of gravitation where you will need to look up the mass of the sun M and the gravitational constant G.

Try that and see how close the answer is.
 
Last edited:
pumpernickel said:
And did F= (V^2)/r and came up with .000855

Aren't you omitting m in that equation?

I would agree with paisiello in that you would need to know the mass of the sun, and apply that to Newtons law.

g = GM/r^2
 
I get a different number of seconds for the period of revolution (somehow you got your period in seconds a factor 2.5 too high). If you correct that and write "a" instead of "F" in your last expression (since F/m = a), then you should be good.

(You don't need to use Newtons law of gravitation).
 
Doesn't that assume the asteroid is moving uniformly in a perfect circle? That assumption needs to be stated somewhere.

Otherwise you'll need to use Newton's law.
 
paisiello2 said:
Doesn't that assume the asteroid is moving uniformly in a perfect circle? That assumption needs to be stated somewhere.

The topic of this thread says "Circular motion" and the poster used an equation for centripetal force in a uniform circular motion, so I guess we can safely assume that this is an exercise on the topic of circular motion. Granted, it is always a good habit and it never hurts to state your assumption in case they are not obvious to the reader (or, in this case, the teacher checking your work).
 

Similar threads

Replies
335
Views
17K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K