How Strong Is the Sun's Gravity at Earth's Orbit Distance?

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Homework Help Overview

The discussion revolves around calculating the gravitational acceleration of the Sun at the distance of Earth's orbit. Participants are exploring the application of Newton's Law of Universal Gravitation in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the gravitational pull of the Sun as a centripetal force and consider using Newton's law of gravity. There are multiple inquiries about the correct application of the formula and the values for mass and distance.

Discussion Status

Some participants have provided values for the mass of the Sun and the distance to Earth, while others are questioning the correctness of their calculations. There is an ongoing exploration of different approaches to the problem, including using orbital speed.

Contextual Notes

Several posts indicate a lack of clarity on how to begin the calculations, and there are repeated requests for help. The discussion reflects uncertainty about the application of gravitational concepts and the necessary values to use.

guru
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What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?

i have no idea where to begin.
Help
 
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The gravitational pull of the Sun on the Earth is the centripetal force keeping the Earth from flying off into the great beyond.
 
guru said:
What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?

i have no idea where to begin.
Help

Find the mass of the sun, and the distance between the sun and the earth. Then, think about Newton's Law of Universal Gravitation.
 
This is what i obtained
Sun's Mass: 1.99x10^30 kg,
earth-sun distance: 150x10^6 km.

law of gravitational = Gm/r^2
G=6.67*10^(-11)

((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^12)

is that right?
 
Last edited:
guru said:
What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?

i have no idea where to begin.
Help

Probably the simplest and most direct approach is to use Newton's law of gravity.

Do you know what that is?
 
pervect said:
Probably the simplest and most direct approach is to use Newton's law of gravity.

Do you know what that is?
law of gravity = Gm/r^2
G=6.67*10^(-11)

((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^12)
right?
 
The acceleration due to the Sun's gravity at the location of the Earth is just

[tex]a = \frac {v^2}{R}[/tex]

where v is the speed of the Earth around the Sun and R is the radius of Earth's orbit. You know R and you know the Earth travels a distance of [itex]2\pi R[/itex] in one year so you know its speed too.
 
guru said:
law of gravity = Gm/r^2
G=6.67*10^(-11)

((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^12)
right?

That should be right!
 

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