Calculate net gravitational field on the moon, without knowing its mass?

[joey_oggie]

Homework Statement

The Moon is 3.9*10^5 km from Earth's center and 1.5*10^8 km from the Sun's center. The masses of Earth and the Sun are 6*10^24 and 2*10^30 kg, respectively.

a. The ratio of the gravitational fields due to the Sun and Earth at the center of the moon is:
1) 2300
2) 230
3) 23
4) 2.3 < Correct answer

b. When the Moon is in its third quarter phase, its direction from Earth is a right angles to the Sun-Earth direction. The net gravitational field due to the Sun and Earth at the center of the Moon is?:
1) 6.4*10^-3 m/s^2
2) 6.4*10^-6 m/s^2
3) 6.4*10^-8 m/s^2
4) 6.4*10^-10 m/s^2

Homework Equations

Fg=G(m1*m2)/r^2

The Attempt at a Solution

I attached a picture of the diagram I drew. I tried to resolve the forces in the X-Y Coordinates but I didn't know what to do from there; I still didn't know the mass of the Moon so I couldn't get a pure answer like the ones given in the choices.

 

[D H]

You don't need to know the Moon's mass for any of these problems. If you can't see how to do these without knowing the Moon's mass, make up a value. Call it one kilogram, for example.

 

[joey_oggie]

You don't need to know the Moon's mass for any of these problems. If you can't see how to do these without knowing the Moon's mass, make up a value. Call it one kilogram, for example.

I must be doing something wrong because it gave me an answer not in the choices. I'm going to attach a photo (sorry about the quality I took it using my phone) of what I did. I'm not sure about the resolution part, there might be a mistake(s) there. Can you please take a loot it it and tell me where I went wrong? :)

Forgot to add, I'm talking about the 2nd requirement b, not a. (the one with the net gravitational field and 3rd phase). I didn't need the Moon's mass in the first part of the problem.

 

[D H]

You don't need the Moon's mass for the 2nd part of the problem, either. Look at the units on the choices. They are accelerations, not forces. Now look at Newton's second law and the law of gravitation. You can combine these two formulae to yield a formula for the acceleration of a subject body that is independent of the subject body's mass.

The question is asking for acceleration: A vector. How do you add vectors? You can't just add them like scalars, which is what you did.

 

[joey_oggie]

You don't need the Moon's mass for the 2nd part of the problem, either. Look at the units on the choices. They are accelerations, not forces. Now look at Newton's second law and the law of gravitation. You can combine these two formulae to yield a formula for the acceleration of a subject body that is independent of the subject body's mass.

The question is asking for acceleration: A vector. How do you add vectors? You can't just add them like scalars, which is what you did.

Ohhhh yeah, never noticed that part. So that part is understood, thanks!


About the resolution of the forces (or accelerations now) Fa and Fb, do I have to use the following formula?

R^2=X^2 + Y^2 (where R is the net gravitational force, X is the sum of components on X-axis, and Y is the sum of components on the Y-axis)

If yes, then I tried doing it now and I had no problems getting the value of Y but when calculating the sum of forces on the X-axis I couldn't get the Sin(theta) as I don't have the distance between Earth and the Sun. Can I use the calculator to get an approximate value of Theta using the Cos I already have?

 

[SammyS]

Ohhhh yeah, never noticed that part. So that part is understood, thanks!

About the resolution of the forces (or accelerations now) Fa and Fb, do I have to use the following formula?

R^2=X^2 + Y^2 (where R is the net gravitational force, X is the sum of components on X-axis, and Y is the sum of components on the Y-axis)

If yes, then I tried doing it now and I had no problems getting the value of Y but when calculating the sum of forces on the X-axis I couldn't get the Sin(theta) as I don't have the distance between Earth and the Sun. Can I use the calculator to get an approximate value of Theta using the Cos I already have?

Your 1st figure:

Using trig you will find that θ ≈ 89.851°

You can pretty well approximate it as a right angle.

If you use the Pythagorean Theorem to find the Sun to Earth distance, it is only about 507 km less than the Sun to Moon distance of 1.5×10⁸ km of the hypotenuse.