Criteria for Determining Satellite's Return from Orbit?

[Clara Chung]

Homework Statement

Homework Equations

Conservation of energy/ angular momentum

The Attempt at a Solution

I used conservation of momentum to do part d. My answer is V_B/V_A=sinα /X. i don't know how to do part e. What is the criteria to determine whether the satellite will return to the planet's surface?

 

[haruspex]

The question seems strange in at least one regard: the inequality to be proved can be greatly simplified... but you do have to be careful with signs.

It gets even stranger when you figure out the range of X values for which the given inequality is false.

Now, I'm no expert in this area, but I always thought that if an object is a satellite (i.e. below escape velocity) then it follows an elliptical path forever. It cannot start on some other path and migrate to an elliptical orbit without some impulse being applied. So if it starts off rising from the surface at some angle above the horizontal, and is below escape velocity, then...

 

[Clara Chung]

I don't know too. I always thought that it either fire the satellite tangentially or fire the satellite with an angle but it will always return to the earth...

 

[PeroK]

I don't know too. I always thought that it either fire the satellite tangentially or fire the satellite with an angle but it will always return to the earth...

$\frac{X(X-1)}{X^2 - \sin^2\alpha} \le \frac{X(X-1)}{X^2 - 1} = \frac{X}{X + 1} < 1 \ \$ (for $X > 1$)

 

[haruspex]

$\frac{X(X-1)}{X^2 - \sin^2\alpha} \le \frac{X(X-1)}{X^2 - 1} = \frac{X}{X + 1} < 1 \ \$ (for $X > 1$)

I think it is more helpful to simplify the original inequality, working the cases of positive and negative denominator separately.

 

[Clara Chung]

I think it is more helpful to simplify the original inequality, working the cases of positive and negative denominator separately.

How about assume that the velocity when the satellite reach the Earth be v.
By conservation of energy,
mv_A² / 2 = mv² /2
=> v_A=v
By conservation of angular momentum,
v_A sin α R = v R
=> sin α = 1 ?

 

[PeroK]

How about assume that the velocity when the satellite reach the Earth be v.
By conservation of energy,
mv_A² / 2 = mv² /2
=> v_A=v
By conservation of angular momentum,
v_A sin α R = v R
=> sin α = 1 ?

I wouldn't worry about this problem. There's definitely something not quite right.

 

[haruspex]

I don't know too. I always thought that it either fire the satellite tangentially or fire the satellite with an angle but it will always return to the earth...

Not if it is fired fast enough (escape velocity), but then B would be at infinity, making X infinite.

For what little it is worth, the simplification of the given inequality goes like this:
If the denominator > 0 (i.e. X>sin(α)) then we can multiply out to get
X(X-1)<X²-sin²(α)
Cancelling an X²:
-X<-sin²(α)
X>sin²(α)
And since we are assuming X>sin(α) that is trivially true.
But if X<sin(α) then when we multiply out to there is a reversal:
X(X-1)>X²-sin²(α)
Cancelling an X²:
-X>-sin²(α)
X<sin²(α)
So the given inequality is equivalent to:
X<sin²(α) or X>sin(α).

The implication, sort of, is that if sin²(α) < X<sin(α) then it will not crash back to the planet. This is clearly absurd. If X<1 then the point B is inside the planet.

 

[PeroK]

The implication, sort of, is that if sin²(α) < X<sin(α) then it will not crash back to the planet. This is clearly absurd. If X<1 then the point B is inside the planet.

$X > 1$ by its definition in the problem.