Gravitational Forces: Ratio of Earth & Sun on Moon

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Homework Help Overview

The problem involves calculating the ratio of gravitational forces exerted by Earth and the Sun on the Moon, given specific distances and masses. The context is within gravitational physics, focusing on the application of Newton's law of universal gravitation.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore different methods for calculating the gravitational force ratio, including direct calculations and simplified ratio expressions. Some question the precision of their calculations and whether they are considering all necessary factors.

Discussion Status

There are multiple interpretations of the problem, with participants attempting various calculations and expressing uncertainty about the correctness of their results. Some suggest that the focus should be on ratios rather than absolute values, while others question the assumptions made in their approaches.

Contextual Notes

Participants note potential issues with precision in calculations and the relevance of including certain constants in their ratio expressions. There is also mention of duplicate threads, indicating ongoing discussion and attempts to clarify the problem.

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Homework Statement


The moon is 3.9 × 105 km from Earth's center and 1.5 × 108 km from the sun's center. If the masses of the moon, Earth, and sun are 7.3 × 1022 kg, 6.0 × 1024 kg, and 2.0 × 1030 kg, respectively, find the ratio of the gravitational forces exerted by Earth and the sun on the moon. (Use G = 6.670 × 10-11 Nm2/kg2.)

Homework Equations


F=G(m1m2)/d²
Where F is the force, G is gravity constant, m1 and m2 are masses, and d is distance.

The Attempt at a Solution


F=G(mEarthmMoon)/d²
F=6.67×10-11((6×1024)(7.3×1022))/3.9×108²
F=192074950690335305719.92110453649
F=1.9e+20N

F=G(mSunmMoon)/d²
F=6.67×10-11((2×1030)(7.3×1022))/1.5×1011²
F=432808888888888888888.88888888889
F=4.3e+20N

1.9e+20N / 4.3e+20 = 2.2
and I tried its inverse, 0.44

Neither of these are correct.
I am probably missing something from the question.
What is wrong about this and where?
 
Last edited:
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312213 said:

Homework Statement


The moon is 3.9 × 105 km from Earth's center and 1.5 × 108 km from the sun's center. If the masses of the moon, Earth, and sun are 7.3 × 1022 kg, 6.0 × 1024 kg, and 2.0 × 1030 kg, respectively, find the ratio of the gravitational forces exerted by Earth and the sun on the moon. (Use G = 6.670 × 10-11 Nm2/kg2.)

Homework Equations


F=G(m1m2)/d²
Where F is the force, G is gravity constant, m1 and m2 are masses, and d is distance.

The Attempt at a Solution


F=G(mEarthmMoon)/d²
F=6.67×10-11((6×1024)(7.3×1022))/3.9×108²
F=192074950690335305719.92110453649
F=1.9e+20N

F=G(mSunmMoon)/d²
F=6.67×10-11((2×1030)(7.3×1022))/1.5×1011²
F=432808888888888888888.88888888889
F=4.3e+20N

1.9e+20N / 4.3e+20 = 2.2
and I tried its inverse, 0.44

Neither of these are correct.
I am probably missing something from the question.
What is wrong about this and where?

First of all lose the precision. It only makes it difficult to see what's going on.

Second of all since they want a ratio. Take it right upfront and recognize that you can express it more simply as:

Fe/Fs = Me*ds2/Ms*de2

Makes calculation a little easier.
 
Me*ds2/Ms*de2
(6×1024*1.5×1011²)/(2×1030*3.9×108²)
0.44

Imputing this as an answer results in an incorrect.

I'm not really sure what the question is asking for, since what I did seems to be correct, as far as I can tell.
 
The moon is 3.9 × 105 km from Earth's center {Re} and 1.5 × 108 km {Rs} from the sun's center. If the masses of the moon, Earth, and sun are 7.3 × 1022 kg, 6.0 × 1024 kg {Me}, and 2.0 × 1030 kg {Ms}, respectively, find the ratio of the gravitational forces exerted by Earth and the sun on the moon. (Use G = 6.670 × 10-11 Nm2/kg2.)

Earth to Sun Ratio = (6.0 x 1024 kg) * (1.5 × 108 km)2 / 2.0 × 1030 kg * (3.9 × 105 km)2 = .4438

Sun to Earth ratio = 2.253

The numbers look right.
 
Hi 312213! :wink:

Well, that method should work …

but why are you putting G and mmoon into the figures?

This is a dimensions problem, and all you need is the ratios.

Try again! :smile:
 
Well this was just the first way I took it. As ratios, the numbers are the same.

Fe/Fs = Me*ds2/Ms*de2

Me*ds2/Ms*de2
(6×1024*1.5×1011²)/(2×1030*3.9×108²)
0.44

Still 0.44.
I think the question is looking at something else. Or wrong altogether.
 
Merged two duplicate threads into this one...
 

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