Satellite fall and friction force problem

In summary, the 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 a spiral and fall to the Earth. However, if we assume that the friction force is small, the orbit would remain circular. The change in radius at every revolution can be calculated using the equation Δr = \frac{2r\sqrt{r}f}{m_s\sqrt{GM_e}}. The time for the satellite to complete one lap can also be determined using the equation Δt = \frac{2πr\sqrt{r}}{\sqrt{GM_e}}.
  • #1
Sergio Rodriguez
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4

Homework Statement


A satellite of mass [itex]m_s[/itex] orbits the Earth in a circular orbit of radius [itex]r_0[/itex]. 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:
[itex] ΔME = \frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r-Δr)} [/itex]
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: [itex] Δr = \frac {4πr^2}{4πrf - 2Gm_sM_t}[/itex] very different from the the solution of the book: [itex]Δr = \frac {fr^{\frac {3} {2} } }{m_s\sqrt{M_tG} }[/itex]

I have spent sveral hours with this exercise and don't know where is the error. Plese, help.:headbang:
 
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  • #2
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?
 
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  • #3
Sergio Rodriguez said:
The change in mechanical energy:
What about KE?
 
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  • #4
Thanks! I'll check it.
 
  • #5
haruspex said:
What about KE?

That expression of ME already has the KE inside.
 
  • #6
Sergio Rodriguez said:
That expression of ME already has the KE inside.
Ah, yes.. the 2 in the denominator.
Sergio Rodriguez said:
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.
 
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  • #7
Thanks, haruspex! I ' ll post them. And also check the units.
 
  • #8
Sergio Rodriguez said:
Thanks, haruspex! I ' ll post them. And also check the units.
Please see the last sentence I added to my post above.
 
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  • #9
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!
 
  • #10
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

[itex]2πrf = \frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r - Δr} [/itex]
[itex]2r(r - Δr) 2πrf = - (r-Δr) Gm_sM_e + rGm_sM_e[/itex]
[itex]2πrf[2r^2-2rΔr] = - rGm_sM_e+ΔrGm_sM_e + rGm_sM_t[/itex]
[itex] 4πr^3f-4πr^2Δrf = ΔrGm_sM_t[/itex]
[itex]4πr^3f = ΔrGm_sM_t + 4πr^2Δrf [/itex]
[itex]4πr^3f = Δr(Gm_sM_t + 4πr^2f [/itex]
[itex] Δr =\frac {4πr^3f} {Gm_sM_t + 4πr^2f} [/itex]
 
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  • #11
Sergio Rodriguez said:
I have corrected my expression and now is dimensionally correct. The units is metres, the meters the radius decrease after each lap.

[itex]2πrf = \frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r - Δr} [/itex]
[itex]2r(r - Δr) 2πrf = - (r-Δr) Gm_sM_e + rGm_sM_e[/itex]
[itex]2πrf[2r^2-2rΔr] = - rGm_sM_e+ΔrGm_sM_e + rGm_sM_t[/itex]
[itex] 4πr^3f-4πr^2Δrf = ΔrGm_sM_t[/itex]
[itex]4πr^3f = ΔrGm_sM_t + 4πr^2Δrf [/itex]
[itex]4πr^3f = Δr(Gm_sM_t + 4πr^2f [/itex]
[itex] Δr =\frac {4πr^3f} {Gm_sM_t + 4πr^2f} [/itex]
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.
 
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  • #12
Finally I got it!
To simplify my expression I try the two ways you said me.
First way:

[itex] Δr = \frac {4πr^3f}{Gm_sM_e + 4πfr^2}[/itex]

[itex]Δr = \frac {\frac {4πr^3f}{r^2}}{\frac{Gm_sM_e}{r^2} + \frac {4πfr^2}{r^2}}[/itex]

[itex]Δr = \frac {4πrf}{\frac{Gm_sM_e}{r^2} + 4πf}[/itex]

[itex]Δr = \frac {4πrf}{GF + 4πf}[/itex]

and considering GF >> f

[itex]Δr = \frac {4πr^3f}{Gm_sM_e}[/itex]

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:

[itex] \frac{1}{1-x} = 1 + x^2 + x^3 + ... + x^n [/itex]

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

and using only:
[itex] \frac {1}{1 - \frac{Δr}{r}} ≈ 1 + \frac {Δr}{r}[/itex]

I got:

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

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

The time the satellite need to do one lap:

[itex]v = \sqrt{\frac{GM_e}{r}}[/itex]

[itex]\frac{2πr}{Δt} = \sqrt{\frac{GM_e}{r}} [/itex]

[itex]Δt = \frac{2πr\sqrt{r}}{\sqrt{GM_e}}[/itex]

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

[itex]\frac{Δr}{Δt} = \frac{4πr^2f\sqrt{GM_e}}{Gm_sM_e2πr\sqrt{r}}[/itex]

[itex]\frac {Δr}{Δt} = \frac{2r^2f\sqrt{GM_e}}{Gm_sM_e\sqrt{r}} ⋅ \frac{\sqrt{GM_er}}{\sqrt{GM_er}}[/itex]

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

Thank you very much you two!:partytime:
 

Related to Satellite fall and friction force problem

1. What is satellite fall?

Satellite fall is the process of a satellite descending towards the Earth's surface due to the gravitational force acting upon it.

2. What causes a satellite to fall?

A satellite falls due to the Earth's gravitational force pulling it towards the surface. This force increases as the satellite gets closer to the Earth's surface.

3. How does friction affect a satellite's fall?

Friction can slow down the speed at which a satellite falls by creating drag, which is a force that opposes the motion of the satellite. However, friction also generates heat, which can cause a satellite to burn up as it falls through the Earth's atmosphere.

4. What factors affect the amount of friction a satellite experiences during its fall?

The amount of friction a satellite experiences during its fall can be affected by factors such as the satellite's shape, speed, and the density of the atmosphere it is falling through.

5. Can friction be used to control a satellite's fall?

Yes, friction can be used to control a satellite's fall by adjusting the satellite's shape or using materials that can withstand high levels of heat generated by friction. Additionally, parachutes can be deployed to increase the amount of drag and slow down the satellite's descent.

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