Find the Mass of Sun if given are T and d?

[arddi2007]

Homework Statement

Knowing that the Earth spins around the Sun for 365 days and the distance Earth->Sun is 1.5x10^11 m, what's the mass of the Sun?

Homework Equations

F=G Me x Ms/distance^2

where Me - Mass of Earth, and Ms - Mass of Sun

In some literature, I noticed that for similar problems was used the formula for centripetal acceleration, although I'm not sure why:

acp = v^2/r=(2 x PI x r/T)^2 / r

where r - radius of Earth, and T - time for the Earth to spin around Sun (?).

The Attempt at a Solution

You can notice those two formulas. The literature I found this equation in says that F and acp are equal, but it doesn't explain why. If that's the case, then the equation will be a piece of cake. But I need to know why F (pulling force between Earth and Sun) and acp are equal (if that's correct).

THANK YOU FOR ANY HELP AT ALL!

 

[gneill]

You can notice those two formulas. The literature I found this equation in says that F and acp are equal, but it doesn't explain why. If that's the case, then the equation will be a piece of cake. But I need to know why F (pulling force between Earth and Sun) and acp are equal (if that's correct).

They are not equal, and for a very good reason: one is a force and the other is an acceleration. They are different things.

However! They can be closely related through Newton's second law if the acceleration in question is the result of the force in question acting upon a given mass. In this case, that mass is the Earth.

You know that gravitational force is holding the Earth in its orbit around he Sun. This force provides the centripetal acceleration of the Earth in its orbit. Can you find an expression for the gravitational acceleration of the Earth by the Sun?

 

[arddi2007]

Now that I've checked the literature again, it says that the centripetal force of Earth (Fcp=mv^2 / distance) is equal to the force which Earth and Sun pull each other.

Now, my question is, is that correct? If it is, why are they equal?

 

[gneill]

Now that I've checked the literature again, it says that the centripetal force of Earth (Fcp=mv^2 / distance) is equal to the force which Earth and Sun pull each other.

Now, my question is, is that correct? If it is, why are they equal?

That is correct, assuming a circular orbit for the Earth.

In reality, Earth's orbit is not perfectly circular but very slightly elliptical. So there are small variations in the distance between the Earth and Sun over the course of a year that affects the gravitational force between them. The velocity of the Earth in its orbit also varies slightly over the year as it follows the ellipse. But none of this is going to affect what it is you want to do here.

Why not start by calculating a number for the centripetal acceleration?

 

[rwisz]

I would first clarify the units used in the given values. The time, T, is in years, while the distance, d, is in meters. This means that the value for T needs to be converted to seconds for the equations to work properly. Additionally, it would be useful to convert the distance to kilometers, as the mass of the Sun is typically measured in kilograms and the gravitational constant, G, is typically given in units of m^3/kg*s^2.

Once this is done, we can use the first equation listed to solve for the mass of the Sun. This equation is derived from Newton's law of gravitation, which states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. By setting the force of gravity between the Earth and Sun equal to the centripetal force keeping the Earth in orbit, we can solve for the mass of the Sun.

In order to use the second equation listed, we would need to know the radius of the Earth's orbit, not the radius of the Earth itself. This can be calculated using the given distance and the fact that the Earth's orbit is approximately circular. However, this equation may not be as accurate as using the first equation, as it does not take into account the mass of the Earth. It also assumes that the orbit is perfectly circular, which is not entirely accurate.

In conclusion, the first equation using Newton's law of gravitation would be the most accurate and appropriate way to solve for the mass of the Sun given the information provided. However, it is important to clarify the units and make any necessary conversions before using the equation.