How to calculate Earth speed of the Moon induced orbit?

[emily-]

Homework Statement:

How to calculate earth speed of the moon induced orbit?

Relevant Equations:

I thought of using F = ma along with momentum p =mv to get the v for earth

It didn't work and I don't know how to do it.

 

[Drakkith]

Sorry, what is an 'induced' orbit? A quick google search didn't give me any results for that term.

 

[jbriggs444]

Homework Statement:: How to calculate earth speed of the moon induced orbit?
Relevant Equations:: I thought of using F = ma along with momentum p =mv to get the v for earth

It didn't work and I don't know how to do it.

Can you clarify the question? Are you trying to calculate the speed of the moon in its orbit about the Earth? Or the Earth in its orbit about the Earth-moon barycenter? The speed of the Earth's tidal bulge as it "orbits" the Earth's surface?

What inputs do you have to play with?

Since the only work you have shown is a pair of equations that "didn't work", I'll leave it there. We need you to show more work than that.

 

[emily-]

Sorry, what is an 'induced' orbit? A quick google search didn't give me any results for that term.

Can you clarify the question? Are you trying to calculate the speed of the moon in its orbit about the Earth? Or the Earth in its orbit about the Earth-moon barycenter? The speed of the Earth's tidal bulge as it "orbits" the Earth's surface?

What inputs do you have to play with?

Since the only work you have shown is a pair of equations that "didn't work", I'll leave it there. We need you to show more work than that.

I am thinking about the speed that earth has to have due to its gravitational interaction with the moon. I know that the moon's orbital velocity is approximately 1.022 km/s
Now according to Newton's third law, the force is equal for both of the bodies meaning that F = m_earth * a_earth = m_moon * a_moon
m_earth, m_moon and a_moon are known. By replacing the variables with values we get the value for a_earth. Now according to this formula F * dt = p = mv: a_earth * m_earth * dt = m_earth * v_earth which mean that
v_earth = a_earth * dt

 

[jbriggs444]

I am thinking about the speed that earth has to have due to its gravitational interaction with the moon. I know that the moon's orbital velocity is approximately 1.022 km/s

OK. So it is the Earth's orbital speed about the Earth-moon barycenter that you are hoping to calculate.

Now according to Newton's third law, the force is equal for both of the bodies meaning that F = m_earth * a_earth = m_moon * a_moon
m_earth, m_moon and a_moon are known. By replacing the variables with values we get the value for a_earth.

Yes indeed. If we knew $a_\text{moon}$ we could solve for $a_\text{earth}$. Though we do have the pesky problem of finding $a_\text{moon}$ if we attack the problem that way.

Now according to this formula F * dt = p = mv:

That is not a correct formula. $F\ dt = dp$. Force is the rate of transfer of momentum. But that rate is an incremental change, not a final total. It is not equal to $p$ and hence, is not equal to $mv$.

You'd need to integrate $F \sin \theta\ dt$ over one quarter cycle to get one component of the momentum. Not that bad as integrals go. But there are easier approaches. [Exercise for you: Justify the calculation of momentum using that integral]

You may want to think about formulas for centrifugal force or centripetal acceleration. Or about the location of the center of mass of the system.