What is the Force Exerted by the Sun on Earth?

  • Thread starter Thread starter tag16
  • Start date Start date
  • Tags Tags
    Force Magnitude
Click For Summary
The discussion focuses on calculating the gravitational force exerted by the Sun on Earth using two methods: Newton's law of universal gravitation and centripetal force. The first method yields a force of approximately 36.75 x 10^21 N, while the second method gives a slightly lower value of 32.40 x 10^21 N. The discrepancy between the two results is attributed to the assumptions made regarding the Sun's mass and the Earth's elliptical orbit. Both approaches highlight the importance of accurate data and the nature of the Earth's orbit in determining gravitational force. Understanding these calculations is essential for grasping the dynamics of celestial bodies.
tag16
Messages
95
Reaction score
0

Homework Statement


The orbital speed of Earth about the Sun is 2.8x10^4 m/s and its distance from the Sun is 1.5x10^11 m. The mass of Earth is approximately 6.2x10^24 kg and that of the Sun is 2.0x10^30 kg. What is the magnitude of the force exerted by the Sun on Earth?


Homework Equations


F= Gm1m2/r^2 and F=mv^2/r


The Attempt at a Solution


I know I need to use Newtons 2nd Law but every time I plug the numbers in I get the wrong answer. I tried using F= Gm1m2/r^2 and F=mv^2/r
 
Physics news on Phys.org
What's the text-book answer that's making you think you're doing it wrong?
You can solve this question in two different ways, and for both, there would be redundant data.

Well, since you were already on the right track already, I'll just show you what I got when I plugged the numbers in myself.

If you know that the force the sun exerts on the Earth is gravity, then you can simply use:

|F| = GMEMS/R² = (6.67*6.2*2)/1.5² * 10^(-11+24+30-22) = 36.75*10^21 N

If you don't know that the force is gravity, however, but do know that the Earth revolves around the sun is an approximately circular orbit, you can say that the magnitude of the force the sun exerts upon it is:
|F| = ME*V²/R = 6.2*2.8²/1.5 * 10^(24+8-11) = 32.40*10^21 N

We need to explain the discrepancy here somehow, though. I'd chalk it up to poor approximations of the sun's mass (Note how the centripetal force equation does not directly use the data about the mass of the sun) and the fact that the Earth orbits the sun in an elliptic orbit, and not a circular one.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
5K
Replies
1
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
Replies
2
Views
1K
Replies
13
Views
1K
Replies
11
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K