Gravitational Force Problem - Help

AI Thread Summary
The discussion revolves around calculating the ratio of maximum to minimum distance (rmax/rmin) for a moon orbiting a planet, given that the maximum gravitational force exceeds the minimum by 11%. The initial calculation yielded a ratio of 0.949, while the textbook states it should be 1.05, leading to confusion about the definitions of rmax and rmin. Participants clarified that rmin refers to the closest distance where gravitational force is greatest, resolving the ambiguity in the problem's wording. The hint provided helped participants understand the correct interpretation, allowing them to solve the problem accurately. The conversation highlights the importance of clear definitions in physics problems.
SpacemanRich
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Homework Statement


As a moon follows it's orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minimum gravitational force by 11%. Find the ratio rmax / rmin, where rmax is the moon's maximum distance from the center of the planet and rmin is the minimum distance.

Homework Equations


Newton's law of Gravitation F = G m1m2 / r2

The Attempt at a Solution


I found the ratio to be, rmax / rmin = 0.949 to 3 significant digits.
My textbook says the ratio is 1.05. I think the text if wrong. Can someone try this problem and let me know what you get.
Thanks,
R.
 
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SpacemanRich said:

Homework Statement


As a moon follows it's orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minimum gravitational force by 11%. Find the ratio rmax / rmin, where rmax is the moon's maximum distance from the center of the planet and rmin is the minimum distance.

Homework Equations


Newton's law of Gravitation F = G m1m2 / r2

The Attempt at a Solution


I found the ratio to be, rmax / rmin = 0.949 to 3 significant digits.
My textbook says the ratio is 1.05. I think the text if wrong. Can someone try this problem and let me know what you get.
Thanks,
R.
If rmax / rmin < 1, then rmax < rmin .

Does that make any sense?

Where is the gravitational force greater, at rmin or at rmax?
 
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SpacemanRich said:

Homework Statement


As a moon follows it's orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minimum gravitational force by 11%. Find the ratio rmax / rmin, where rmax is the moon's maximum distance from the center of the planet and rmin is the minimum distance.

Homework Equations


Newton's law of Gravitation F = G m1m2 / r2

The Attempt at a Solution


I found the ratio to be, rmax / rmin = 0.949 to 3 significant digits.
My textbook says the ratio is 1.05. I think the text if wrong. Can someone try this problem and let me know what you get.
Thanks,
R.

At first I thought you were right, but there is an ambiguity in the notation. Is rmax the radius where the force is max, or is it the maximum radius? You get the two different answers depending on how you define the subscripts of the radius variable...
 
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Hint

F_{max} = \frac{Gm_1m_2}{r_{min}^2}
 
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SammyS said:
If rmax / rmin < 1, then rmax < rmin .

Does that make any sense?

Where is the gravitational force greater, at rmin or at rmax?

The gravitational force is greatest at rmin.
The confusion was over whether rmin meant the point where the moon is closest to the planet, or if it meant the variable r of the Gravitational formula for Fmin. From the responses I got, I was not the only one confused by the wording of the problem.
 
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berkeman said:
At first I thought you were right, but there is an ambiguity in the notation. Is rmax the radius where the force is max, or is it the maximum radius? You get the two different answers depending on how you define the subscripts of the radius variable...

I'm glad I was not the only one confused by the wording of the problem. The hint given by kreil clarified it, and the problem works out fine then. Thanks for your input.
 
kreil said:
Hint

F_{max} = \frac{Gm_1m_2}{r_{min}^2}

That makes it perfectly clear what I was missing when reading the problem. Now the problem works as expected.

Thanks for opening my eyes..

Regards,
Rich
 
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