Gravitation Force Using Differentials

[BitterSuites]

[SOLVED] Gravitation Force Using Differentials

Homework Statement

Three part question:

1. Consider a solar system similar to our Sun and Earth, where the mass and radius of the planet are 4.22e24 kg and 6.63e6 m, respectively, the mass of the sun is 2.08e30 kg and the planet-sun distance is 1.423e11 m.

Find the magnitude of the change in F in gravitational force that the Sun exerts on a 50.1kg woman standing on the equator at noon and midnight. Assume the Sun and the planet are the only masses acting on the woman. Since change in r is so small, use differentials. Answer in units of N.

2. Find the magnitude of the factional percent change theta F/F % in the Sun's gravitational force on the woman due to the rotation of the Earth in the 12 hours between noon and midnight. Assume the SUn and the Earth are the only masses acting on the woman. Answer in units of %.

3. If the moon orbiting the planet has mass 7.26e22 kg and the planet-moon distance is 3.71e8 m, find the magnitude of the fractional percent change in the moon's gravitational force on the woman due to the rotation of the Earth in the 12 hours between noon and midnight. Assume the Earth and the Moon are the only masses acting on the woman. Answer in units of %.

Homework Equations

Force of the sun on the woman at noon - Force of the sun on the woman at midnight.

The Attempt at a Solution

F of the sun on the woman at noon = Gm1m2/r^2 = ((6.67 x 10^-11)(50.1)(2.08 x 10^30))/(1.423 x 10^11)^2 = .343255

F of the sun on the woman at midnight = Gm1m2/r^2

Since the woman is now 6.63 x 10^6 m further away from the sun, the new r is 1.4230663 x 10^11

So F = ((6.67 x 10^-11)(50.1)(2.08 x 10^30))/(1.4230663 x 10^11)^2 = .343223

So the difference is .343255 - .343223 = .000032 N

However, this answer is incorrect.

Question 2 will simply be the change in force divided by the force, but I can't answer until I figure out #1. Question 3 should just be repeating the same process used for #1 and #2.

What is the flaw in my thinking? Any help is appreciated.

 

[Saladsamurai]

If you were to draw a free-body diagram of the WOMAN, how many forces should be acting on her at any given time?

 

[BitterSuites]

It would be gravity from Earth and the gravitational constant from the Sun, correct?

 

[Saladsamurai]

I am not sure what you mean by the gravitational constant from the sun. Could you be a little more specific?

 

[BitterSuites]

I get the feeling that you're saying this is where I'm making my mistake, but what I'm referring to is 6.67 x 10^-11 N m^2/kg^2. In other words, big G instead of little g.

 

[Gear300]

What you did seems to be alright...dont see much wrong with it; but the concept of differentials, however, is another method: If f(x) has derivative dy/dx, then dy/dx = f'(x)... If you were to take dy/dx to be the slope of the tangent line, then dy = f'(x)*dx; meaning the increment in f(x) is the derivative multiplied with the increment in x. To apply that here, you would find the derivative of the gravitational force and multiply it with the increment in x (the additional diameter); it'll get you the same answer. Anyways...your actual problem is that you're using the radius instead of the diameter of the earth.

 

[Saladsamurai]

Okay. But G is not a force. It is merely a constant. It is not exclusive to the sun either. G is G every where in the universe (as far as we know).

Now. Here is my point. You actually need two free body diagrams. One showing the forces and direction of a woman standing on the equator at midnight and one at noon. What are the differences? You have pointed out one of them: that the planet-sun distance is differentially smaller.

Is there anything else?
Hint: Gravitational force is always a pulling force. . . pulling where?

EDIT: Nevermind this part. I was looking at the NET gravitational force. I did not see that the question was only asking about the sun's gravitational force only!

 

[BitterSuites]

I think I may be hopelessly confused :D

Gear300: I tried switching the two values and it was not correct.

Salad: Gravitational force pulls towards the object being acted up, correct? So the forces would be Earth's pull on her and her pull on the Sun?

 

[Gear300]

Sorry BitterSuites...I accidentally missed magnitude of difference. Your problem is you're using radius of Earth not diameter. Noon and Midnight I think are on opposite sides

 

[BitterSuites]

So in the second equation, should I have divided by (1.42303315 x 10^11)^2, as that would be the difference in their radii squared?

 

[Gear300]

If that's the addition of twice the radius, then it should be. Check and see.

 

[BitterSuites]

Twice the radius would just be the diameter, correct? That is what I used originally.

 

[Gear300]

With twice the radius, it should be 1.42313E11. The answer I got for that was a difference of around .000061N...is that the answer?

 

[BitterSuites]

I'm so dumb for doing that.

I plugged that in, but the system still tells me that I am incorrect.

Is there a flaw in my formulation? Have I used incorrect factors?

Am I correct to attempt to find the force of the sun on the woman at midnight minus the force of the sun on the woman at noon.

Those equations looking as I posted them above?

*tries to find his way out of the forest of confusion*

 

[Gear300]

wait...so the one I gave was wrong? If that's the case, then I guess I'll have to rethink.

 

[BitterSuites]

Oh goodness. The system rounded differently from my calculator and didn't have a margin of error.

The correct answer was 6.39962e-05, which is what I got after inputing the correct radius as you pointed out. I had put 6.4e-05 and was counted wrong. Frustrating.

 

[BitterSuites]

For #2, would I use the F at noon or at midnight to determine the percentage change?

I'm thinking it would be from noon, but this professor makes me feel gunshy.

 

[Gear300]

yes...it'd be between noon and midnight

 

[BitterSuites]

Thank you for all of your help. Problem solved.