Help with gravitational problem

[toesockshoe]

Homework Statement

Given that we know the mass of the moon and the Earth and the distance between their centers as the moon orbits the earth, if the Earth's angular velocity about its own axis is slowing down from say some initial given omega to a final omega (due to tidal friction in reality), find out what happens to the orbital distance between the Earth and moon as a consequence

Homework Equations

$F_g = \frac{Gm_mm_E}{r^2}$[/B]

The Attempt at a Solution

My first guess is that this is an angular momentum conservation problem because if you make the origin the center of the earth, there are no net external torques.

soo...
$L_i = L_f$
$r_{imoon} x p_i + I_{earth} \omega _i = r_{fmoon} x p_f + I_{earth} \omega _f$
$r_im_m \frac {v_{imoon}}{r_i} + I\omega_i = r_fm_m \frac{v_{fmoon}}{r_f} + I\omega_f$

obviously this is wrong, as the r's cancel out and we arent left with the variable we are trying to solve. How would I go about solving this problem and where am I going wrong?[/B]

 

[haruspex]

What is x?
Where do the divisors r_i, r_f come from?
You need another equation, one relating the moon's orbital radius to its speed (linear or angular).

 

[toesockshoe]

What is x?
Where do the divisors r_i, r_f come from?
You need another equation, one relating the moon's orbital radius to its speed (linear or angular).

The x is supposed to be a cross... To multiply vectors. Ri is given the distance from Earth to moon, and we are trying to find how Rf changes.

Is the relationship just $v_{earth} + \omega _{moon} = v_{moon}$ ... so $v_{earth} + \frac{v_{moon}}{R} = v_{moon}$ ?

 

[haruspex]

The x is supposed to be a cross... To multiply vectors. Ri is given the distance from Earth to moon, and we are trying to find how Rf changes.

Is the relationship just $v_{earth} + \omega _{moon} = v_{moon}$ ... so $v_{earth} + \frac{v_{moon}}{R} = v_{moon}$ ?

To show a cross product in Latex use \times.
I'm asking how you get r_i and r_f divisors in the last line of your original post. What are you substituting for p_i and p_f?

 

[toesockshoe]

To show a cross product in Latex use \times.
I'm asking how you get r_i and r_f divisors in the last line of your original post. What are you substituting for p_i and p_f?

p=mv... so but the velocity of the moon isn't linear as its rotating around the earth, so i said I tried to figure out how to model the velocity, but i went wrong there.

 

[haruspex]

p=mv... so but the velocity of the moon isn't linear as its rotating around the earth, so i said I tried to figure out how to model the velocity, but i went wrong there.

So do that substitution, correctly this time, and post the equation you get. As I mentioned, you need a second equation to relate the orbital radius to the speed. Think centripetal.

 

[toesockshoe]

So do that substitution, correctly this time, and post the equation you get. As I mentioned, you need a second equation to relate the orbital radius to the speed. Think centripetal.

yeah i wasnt sure, how to do the substitution. i completely lost on this problem. can you explain to me how to proceed?

 

[haruspex]

yeah i wasnt sure, how to do the substitution. i completely lost on this problem. can you explain to me how to proceed?

You have $r_{im} \times p_{im} + I_{e} \omega _{ie} = r_{fm} \times p_{fm} + I_{e} \omega _{fe}$, $p_{im}=m_mv_{im}$, $p_{fm}=m_mv_{fm}$. What's hard about doing the substitutions of the p_*m?
What is the centripetal force required to keep the moon orbiting at radius r? Where is that force coming from? What equation does that allow you to write?

 

[toesockshoe]

You have $r_{im} \times p_{im} + I_{e} \omega _{ie} = r_{fm} \times p_{fm} + I_{e} \omega _{fe}$, $p_{im}=m_mv_{im}$, $p_{fm}=m_mv_{fm}$. What's hard about doing the substitutions of the p_*m?
What is the centripetal force required to keep the moon orbiting at radius r? Where is that force coming from? What equation does that allow you to write?

ok so $bm_{moon}v_{imoon} + I_e \omega _{ie} = bm_{moon}v_{fmoon} + I\omega_ {ef}$

entripetal force: $F_g = m \frac{v^2}{r}$ ... the force is coming from the gravitational force from the earth... so this equation give me the velocity of the moon (because of the v of in the equation).. so i can plug this v value into the v value in the momentum equation... am i correct?

 

[haruspex]

ok so $bm_{moon}v_{imoon} + I_e \omega _{ie} = bm_{moon}v_{fmoon} + I\omega_ {ef}$

What's b? What happened to $r_{im}$ and $r_{fm}$?

so this equation give me the velocity of the moon (because of the v of in the equation)

It gives you two equations, one for each of initial and final radius. Post them.

 

[toesockshoe]

What's b? What happened to $r_{im}$ and $r_{fm}$?

It gives you two equations, one for each of initial and final radius. Post them.

oh my bad... i should have written: r_{im}m_{moon}v_{imoon} + I_e\omega_{ie} = r_[fm}m_{moon}v_f{moon} + I\omega_{ef} [/itex]
so the 2 equations are

$v_i = \sqrt {\frac{Gm_e}{r_i}}$
$v_f = \sqrt {\frac{Gm_e}{r_f}}$ ... right? ... then its a straight plug and chug? and see the relationship as $\omega$ decreases right?

 

[haruspex]

oh my bad... i should have written: r_{im}m_{moon}v_{imoon} + I_e\omega_{ie} = r_[fm}m_{moon}v_f{moon} + I\omega_{ef} [/itex]
so the 2 equations are

$v_i = \sqrt {\frac{Gm_e}{r_i}}$
$v_f = \sqrt {\frac{Gm_e}{r_f}}$ ... right? ... then its a straight plug and chug? and see the relationship as $\omega$ decreases right?

Yes.