Uniform motion earth's orbit question.

[rcmango]

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

before i attempt this problem, can someone help me identify the given information that i may need that is not given in the question, maybe gravity (9.8) or where to use it.

i believe i may need this formula (2pie*r)/T

also may need to find rpm's and convert to rev's, in order to get velocity, and then use the velocity in a formula to get the acceleration. I think that's the approach i may need, its just the information isn't available in the question.

this was from memory. maybe some typos.

thankyou for any help.

 

[OlderDan]

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

before i attempt this problem, can someone help me identify the given information that i may need that is not given in the question, maybe gravity (9.8) or where to use it.

i believe i may need this formula (2pie*r)/T

also may need to find rpm's and convert to rev's, in order to get velocity, and then use the velocity in a formula to get the acceleration. I think that's the approach i may need, its just the information isn't available in the question.

this was from memory. maybe some typos.

thankyou for any help.

You need the law of universal gravitation and Newton's second law, and the values of a few contants (masses, distance, G) and that is all.

 

[kyle8921]

I would like to provide a response to the question about the acceleration due to gravity of the sun at the distance of Earth's orbit. Firstly, it is important to note that the acceleration due to gravity is a constant value of 9.8 m/s^2 on Earth's surface, but it can vary at different distances from the sun. In order to calculate the acceleration due to gravity at Earth's orbit, we need to consider the gravitational force between the sun and Earth.

The formula for gravitational force is F = G(m1m2)/r^2, where G is the universal gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. In this case, m1 would be the mass of the sun and m2 would be the mass of Earth.

The distance between the sun and Earth's orbit is approximately 150 million kilometers. Using this value for r in the formula, we can calculate the gravitational force between the sun and Earth. However, we also need to consider the fact that Earth is in a state of uniform circular motion around the sun, meaning it is constantly changing direction but maintaining a constant speed.

To account for this, we can use the formula for centripetal force, F = mv^2/r, where m is the mass of Earth and v is the velocity of Earth in its orbit. We can calculate the velocity of Earth by dividing the circumference of its orbit (2πr) by the time it takes to complete one orbit, which is one year.

Once we have both the gravitational force and the centripetal force, we can equate them and solve for the acceleration due to gravity at Earth's orbit. This will give us a value of approximately 0.0059 m/s^2, which is significantly lower than the acceleration due to gravity on Earth's surface.

In conclusion, the acceleration due to gravity of the sun at Earth's orbit is approximately 0.0059 m/s^2, which is much smaller than the value of 9.8 m/s^2 on Earth's surface. This is due to the larger distance between the sun and Earth, as well as the fact that Earth is in a state of uniform circular motion around the sun. I hope this helps clarify the approach and information needed to solve this problem.