Satellite fall and friction force problem

[Sergio Rodriguez]

Homework Statement

A satellite of mass $m_s$ orbits the Earth in a circular orbit of radius $r_0$. If the satellite orbits at the upper part of the atmosphere and the friction force f is constant, it would trace an spiral and fall to the earth. but if we suppouse that the friction force is small, so in every moment the orbit would be circular. Calculate the change of the radius at every revolution.

Homework Equations

$$ W_f = ΔME $$
$$ME = \frac {-Gm_sM_e}{2r}$$

The Attempt at a Solution

[/B]
The work done by the friction force is: f ⋅ 2πr, where 2πr the length of the orbit with radius r.
The change in mechanical energy:
$ΔME = \frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r-Δr)}$
because Δr is diference between the radius of one lap and the next one.

$$2πrf =\frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r-Δr)} $$

but when I try to solve for Δr is: $Δr = \frac {4πr^2}{4πrf - 2Gm_sM_t}$ very different from the the solution of the book: $Δr = \frac {fr^{\frac {3} {2} } }{m_s\sqrt{M_tG} }$

I have spent sveral hours with this exercise and don't know where is the error. Plese, help.

 

[gneill]

Check the units in the book's solution. Is the result in meters?

Edit: Maybe they're looking for the rate of change of the radius?

 

[haruspex]

The change in mechanical energy:

What about KE?

 

[Sergio Rodriguez]

Thanks! I'll check it.

 

[Sergio Rodriguez]

What about KE?

That expression of ME already has the KE inside.

 

[haruspex]

That expression of ME already has the KE inside.

Ah, yes.. the 2 in the denominator.

when I try to solve for Δr

I get a different result from your equation. Please post your steps.
You need to use an approximation for small Δr.

But my answer does not match the book's either. Seems to me the book answer is dimensionally wrong, producing a speed, not a distance.

I get the book answer if I find the speed at which the satellite descends, not the distance closer each orbit.

 

[Sergio Rodriguez]

Thanks, haruspex! I ' ll post them. And also check the units.

 

[haruspex]

Thanks, haruspex! I ' ll post them. And also check the units.

Please see the last sentence I added to my post above.

 

[Sergio Rodriguez]

Checking the dimensions of the two expressions, now I see that the book's one is speed units, and mine is dimensionally wrong, so I try to correct it but also try to find the book solution. Thanks!

 

[Sergio Rodriguez]

I have corrected my expression and now is dimensionally correct. The units is metres, the meters the radius has decrease in that lap when the radius is r. Now I try to find dr/dt

$2πrf = \frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r - Δr}$
$2r(r - Δr) 2πrf = - (r-Δr) Gm_sM_e + rGm_sM_e$
$2πrf[2r^2-2rΔr] = - rGm_sM_e+ΔrGm_sM_e + rGm_sM_t$
$4πr^3f-4πr^2Δrf = ΔrGm_sM_t$
$4πr^3f = ΔrGm_sM_t + 4πr^2Δrf$
$4πr^3f = Δr(Gm_sM_t + 4πr^2f$
$Δr =\frac {4πr^3f} {Gm_sM_t + 4πr^2f}$

 

[haruspex]

I have corrected my expression and now is dimensionally correct. The units is metres, the meters the radius decrease after each lap.

$2πrf = \frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r - Δr}$
$2r(r - Δr) 2πrf = - (r-Δr) Gm_sM_e + rGm_sM_e$
$2πrf[2r^2-2rΔr] = - rGm_sM_e+ΔrGm_sM_e + rGm_sM_t$
$4πr^3f-4πr^2Δrf = ΔrGm_sM_t$
$4πr^3f = ΔrGm_sM_t + 4πr^2Δrf$
$4πr^3f = Δr(Gm_sM_t + 4πr^2f$
$Δr =\frac {4πr^3f} {Gm_sM_t + 4πr^2f}$

As I posted, you need to make an approximation for small Δr.
This would have been simplest in the first line. Use the binomial expansion of (1-Δr/r)^-1.
It becomes less obvious what do if done later in your working, but basically you can throw away the second term in the denominator. It will be much smaller than the first term.

 

[Sergio Rodriguez]

Finally I got it!
To simplify my expression I try the two ways you said me.
First way:

$Δr = \frac {4πr^3f}{Gm_sM_e + 4πfr^2}$

$Δr = \frac {\frac {4πr^3f}{r^2}}{\frac{Gm_sM_e}{r^2} + \frac {4πfr^2}{r^2}}$

$Δr = \frac {4πrf}{\frac{Gm_sM_e}{r^2} + 4πf}$

$Δr = \frac {4πrf}{GF + 4πf}$

and considering GF >> f

$Δr = \frac {4πr^3f}{Gm_sM_e}$

Second way:
As I didn't know that binomial expansion ,
I look after it in my Calculus vol 1 (T Apostol) in the section 7.3 p. 339:

$\frac{1}{1-x} = 1 + x^2 + x^3 + ... + x^n$

so in my problem would be:
$\frac {1}{1 - \frac{Δr}{r}} = 1 + \frac {Δr}{r} + (\frac{Δr}{r})^2 + ...$

and using only:
$\frac {1}{1 - \frac{Δr}{r}} ≈ 1 + \frac {Δr}{r}$

I got:

$2πr^2f = -Gm_sM_t + Gm_sM_e(1 + \frac{Δr}{r})$
$2πr^2f = -Gm_sM_t + Gm_sM_e + \frac{Gm_sM_eΔr}{r}$
$Δr = \frac{4πr^3f}{Gm_sM_e}$

The final part, finding Δt was quite easier to do:

The time the satellite need to do one lap:

$v = \sqrt{\frac{GM_e}{r}}$

$\frac{2πr}{Δt} = \sqrt{\frac{GM_e}{r}}$

$Δt = \frac{2πr\sqrt{r}}{\sqrt{GM_e}}$

And dividing Δr and Δt we got: $\frac{Δr}{Δt} = \frac{\frac{4πr^3f}{Gm_sM_e}}{\frac{2πr\sqrt{r}}{\sqrt{GM_e}}}$

$\frac{Δr}{Δt} = \frac{4πr^2f\sqrt{GM_e}}{Gm_sM_e2πr\sqrt{r}}$

$\frac {Δr}{Δt} = \frac{2r^2f\sqrt{GM_e}}{Gm_sM_e\sqrt{r}} ⋅ \frac{\sqrt{GM_er}}{\sqrt{GM_er}}$

$$\frac{Δr}{Δt} = \frac{2r\sqrt{r}f}{m_s\sqrt{GM_e}}$$

Thank you very much you two!