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I was reading through this post today and was very impressed by the response that was given. However, what would have to happen to the velocity in order to collide with the Earth?

Velocity of satellites greater than required velocity

I was think of setting up an equation as follows. If the orbit changes from a circular orbit at some height $h$ with velocity $v$, then an elliptical orbit will occur if the velocity decreases to $\lambda v$, for some $\lambda \in (0,1)$.

From the post made earlier, we know that the original velocity is given by $$v_0^2 = \frac{GM}{R_E+h}$$ and the new velocity is given by $$\lambda^2 v_n^2 = \lambda^2 \Bigg ( GM \Bigg ( \frac{2}{R_E+h} - \frac{1}{a} \Bigg ) \Bigg ).$$ Therefore, solving $$\lambda^2 v_n^2 \leq \frac{GM}{R_E}.$$ Should yield a viable restriction on $\lambda^2$.

But this doesn't give me what I want. A satellite should crash into the earth if it breaks through the atmosphere, i.e when $h < R_E + R_A$, where $R_A$ is the atmospheric height.

How do I determine this $R_A$ from the general theory?

I'm aware that the escape velocity is given by $V_E = \sqrt{\frac{GM}{R_E+h}}$.

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  • $\begingroup$ You can't determine the thickness of the atmosphere from the theory - you have to go and look. It's also a fuzzy boundary: some orbits just crash, but if you just graze the edge of the atmosphere you'll come back out but your orbit will slowly decay and you will crash eventually, though it may take many months. $\endgroup$ – Emilio Pisanty Jul 23 '16 at 11:43
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All you need to do is calculate the perigee distance $r_p$ that is the distance of closest approach. Then if $r_p < R_A$ your satellite will crash and burn.

Once again we start from the vis-viva equation:

$$ v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right) \tag{1} $$

The parameter $a$ is the semi-major axis of the ellipse, and it is related to the perigee and apogee radii as shown below:

Apogee and perigee

So we have:

$$ 2a = r_p + r_a $$

which turns the vis-viva equation (1) into:

$$ v^2 = 2GM\left(\frac{1}{r} - \frac{1}{r_p + r_a} \right) $$

At apogee $r = r_a$ and $v = v_a$ and putting these into our new equation gives:

$$ v_a^2 = 2GM\left(\frac{1}{r_a} - \frac{1}{r_p + r_a} \right) $$

And we just need to rearrange this to get the equation for the perigee distance:

$$ r_p = \frac{r_a}{\frac{2GM}{v_a^{\,2}r_a} - 1} \tag{2} $$

Now let's look at your specific question. We'll call the impact radius $R$, where $R$ would be at least the radius of the Earth but a bit bigger to take into account the atmosphere. So we are looking for the orbit with perigee distance $r_p=R$. The satellite starts in a circular orbit at a radius $r_0$ so the orbital speed is:

$$ v_0 = \sqrt{\frac{GM}{r}} $$

And we ask what happens if we reduce the velocity to $\lambda v_0$. All we have to do is take equation (2) and substitute for the new velocity $v=\lambda v_0$, the apogee radius $r_a=r_0$ and set the perigee radius to the collision radius $r_p=R$ and we get:

$$ R = \frac{r_0}{\frac{2GM}{\lambda v_0^{\,2}r_0} - 1} $$

And on substituting $v_0=\sqrt{GM/r_0}$ this simplifies to:

$$ R = \frac{r_0}{\frac{2}{\lambda} - 1} $$

And rearranging for $\lambda$ gives:

$$ \lambda = \frac{2R}{R+r_0} \tag{3} $$

So given your initial circular orbital radius $r_0$ equation (3) tells you the value of $\lambda$ you need to make your satellite crash and burn.

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  • $\begingroup$ Thanks for the detailed response, however I don't think this quite answers my question. I'm more concerned with the condition on the velocity. So, assuming we start with a circular orbit at a height $R_E+h$ and a velocity $v$. A decrease in the velocity from $v$ to $\lambda v$ will induce an elliptical orbit. Assuming $r_p = R_E + h$ and $r_a = R_E$, what velocities would we compare in order to see how much the velocity has to change in order to have the satellite "crash and burn"? $\endgroup$ – A Bit Too Curious Jul 24 '16 at 0:43
  • $\begingroup$ @Jordan: I've extended my answer to make the calculation clearer $\endgroup$ – John Rennie Jul 24 '16 at 5:19
  • $\begingroup$ Thanks, however, looking through your calculation, are you sure it's not that $$\lambda = \frac{2R_E}{2R_{E}+r_0}?$$ $\endgroup$ – A Bit Too Curious Jul 24 '16 at 11:02
  • $\begingroup$ @Jordan: $R$ is the radius below which the satellite is doomed to crash and burn. If you ignore the atmosphere then $R=R_E$, but in practice $R$ is a bit greater than $R_E$ because you need to include the height of the amosphere. There isn't any simple way to work out exactly what the value of $R$ is because the atmospheric drag depends on the size and shape of the satellite and its velocity in a non-trivial way. $\endgroup$ – John Rennie Jul 24 '16 at 11:28
  • $\begingroup$ @Jordan: where did you get that second factor of $2$ from? It's obviously wrong because if you set $R=r_0$ the equation must give $\lambda=1$. Your equation would give $\lambda=2/3$. $\endgroup$ – John Rennie Jul 24 '16 at 11:30

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