Ratio of the Magnitude of gravitational force?

AI Thread Summary
The discussion revolves around calculating the gravitational forces between a 3.7 kg mass and both the Earth and the Moon, yielding values of 36.37 N and 1.285×10^-4 N, respectively. Participants clarify that the problem also requires calculating the gravitational force between the same mass and the Sun, with the Sun's mass being 1.99×10^30 kg and the distance to it being 1.50×10^8 km. The final goal is to find the ratio of the gravitational force exerted by the Sun to that exerted by the Moon. There is a misunderstanding regarding the units in the ratio, which should be dimensionless as forces cancel out. The correct ratio of the forces is ultimately confirmed to be 171, highlighting the significant difference in gravitational influence.
Albert24
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1. Homework Statement
Calculate the magnitude of the gravitational force between the Earth and a 3.7 kg mass on the surface of the earth. The distance to the center of the Earth from the surface is 6370 km and the mass of the Earth is 5.98·1024.
That gave me 36.37 N

Calculate the magnitude of the gravitational force between the moon and a 3.7 kg mass on the surface of the Earth nearest to the moon. The distance to the center of the moon from the surface of the Earth is 376,000 km and the mass of the moon is 7.36·1022 kg.
That gave me 1.285×10-4 N

Calculate the ratio of the magnitude of the gravitational force between a 3.7 kg mass on the surface of the Earth due to the sun to that due to the moon. The mass of the sun is 1.99·1030 kg and the distance from the center of the sun to the surface of the Earth is 1.50·108 km. ?


2. Homework Equations

F=GMm/r^2

The Attempt at a Solution


I don't quite understand the problem, is it asking what is the magnitude of the gravitonial force between the mass and the sun taking in count the gravitational force between sun anf the moon? In that case the ratio=
GM1m2/r^2/GM1m2/r^2 ??
 
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Albert24 said:
1. Homework Statement
Calculate the magnitude of the gravitational force between the Earth and a 3.7 kg mass on the surface of the earth. The distance to the center of the Earth from the surface is 6370 km and the mass of the Earth is 5.98·1024.
That gave me 36.37 N

Calculate the magnitude of the gravitational force between the moon and a 3.7 kg mass on the surface of the Earth nearest to the moon. The distance to the center of the moon from the surface of the Earth is 376,000 km and the mass of the moon is 7.36·1022 kg.
That gave me 1.285×10-4 N

Calculate the ratio of the magnitude of the gravitational force between a 3.7 kg mass on the surface of the Earth due to the sun to that due to the moon. The mass of the sun is 1.99·1030 kg and the distance from the center of the sun to the surface of the Earth is 1.50·108 km. ?


2. Homework Equations

F=GMm/r^2

The Attempt at a Solution


I don't quite understand the problem, is it asking what is the magnitude of the gravitonial force between the mass and the sun taking in count the gravitational force between sun anf the moon? In that case the ratio=
GM1m2/r^2/GM1m2/r^2 ??
No, the problem is asking what is the gravitational force between the sun and a 3.7 kg mass on the surface of the earth. Call this force Fsun.
You calculated the gravitational force between this same object on the Earth's surface due to the moon, Fmoon, to be 1.285×10-4 N.

What is the ratio Fsun / Fmoon ?
 
Thanks it was 171 N
 
Albert24 said:
Thanks it was 171 N
Can't be.

1. A ratio of two forces has no units.
2. You're saying that the force exerted by the sun on an object sitting on the Earth is orders of magnitude greater than what Earth's gravity exerts on that same object.
Do you think that's reasonable?
 
SteamKing said:
2. You're saying that the force exerted by the sun on an object sitting on the Earth is orders of magnitude greater than what Earth's gravity exerts on that same object.
Do you think that's reasonable?
It's the ratio of the Sun's force to the Moon's force on an object located near the surface of the Earth.
 
I am sorry you are right there are no units they cancel out, since I was working with forces I had Newton in mind :smile:
 
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