Flight Path Angle and Velocity During Atmospheric Re-entry

[AdrianGriff]

A space vehicle enters the sensible atmosphere of the Earth (300,000 ft) with a velocity of 25,000 ft/sec at a flight-path angle of -60 degrees. What is its velocity and flight-path angle at an altitude of 100 nautical miles during descent?

(Assuming no drag or perturbations, two body orbital mechanics)

My answers I am getting are V = 25,370.7 ft/s at a flight-path angle of -60.029 degrees.
The correct answers the book states are: V = 24,618 ft/s at a flight-path angle of 59 degrees and 58 minutes.

There is a clear correlation between my answers and the correct answers, because it seems that my answers are numerically mirrored over the 25,000 ft/s and the 60 degrees.

If someone could work this out mathematically so I can see what I am doing wrong, or just explain if I'm making a simple conceptual mistake, that would be appreciated.

Thank you in advance,
Adrian

 

[magoo]

Why don't you first show us how you came up with your results?

 

[AdrianGriff]

Why don't you first show us how you came up with your results?

Would a photo do the trick?

 

[AdrianGriff]

Why don't you first show us how you came up with your results?

I essentially found the specific angular momentum of the vehicle when it enters the atmosphere (ie 300,000), so h = rv*cos(flight-path angle), and then equated that to h at the altitude of 100 n mi.

Since I need both the veocity and flight path angle at the altitude of 100nmi, I solved for the specific mechanical energy to find v as a function of r, and then found the flight path angle by plugging the velocity I got back into the specific angular momentum equation.

 

[jbriggs444]

I solved for the specific mechanical energy to find v as a function of r

Since your result says that the craft is faster (v=25370 fps) when it is higher (h=100 nautical miles), and since there is no drag, perhaps you could share this part of your work.

Edit to add that... Actual typed-in equations are usually much more easily read than photographed chicken scratchings.

 

[LURCH]

Did you start by standardizing your units of measure? I notice that in the original condition of the spacecraft , the altitude is given in ft, but the question asks for speed and angle at an altitude given in Nautical Miles. What did you get for that? (Just want to make sure you didn’t pull a Mars Climate Orbiter.)

 

[AdrianGriff]

Since your result says that the craft is faster (v=25370 fps) when it is higher (h=100 nautical miles), and since there is no drag, perhaps you could share this part of your work.

Edit to add that... Actual typed-in equations are usually much more easily read than photographed chicken scratchings.

Well, my first instinct to attack this problem is to find both the Specific Mechanical Energy $ξ$ and Specific Angular Momentum $h$. I did this using the velocity $v = 25,000 ft/s$ and $r_{entry} = 300,000ft + 20,902,230ft = 21,202,230 ft$.
$$ξ = \frac {v^2} {2} - \frac {μ} {r}$$
Where $μ = M_⊕*G = 1.408*10^{16} ft^3/s^2$
And
$$h = rv*cos(Φ)$$

Plugging in the values above, I got $ξ = -3.5158 * 10^8 ft/s$ and $h = 2.6503 * 10^{11} ft^2/s$

Now realizing that $ξ$ and $h$ do not change during an orbit, I used the same equations to solve for $v_{instant}$ (i.e. the velocity at the point 100n.mi specified) and $Φ$ the flight-path angle.

Solving using the equation for Specific Angular Momentum $h$, I found that $$v*cos(Φ) = h_{entry}/r_{entry} = 436,182.96 ft/s$$

Then using $ξ$ I solved for the velocity $v_{instant}$:
$$v^2 = 2(ξ + \frac {μ} {r_{instant}})$$
$$v_{instant} = 25,487.07 ft/s$$
And using $v_{instant}$ in the equation for $h$, the flight-path angle is found:
$$Φ=arccos(h/rv)$$

 

[jbriggs444]

Well, my first instinct to attack this problem is to find both the Specific Mechanical Energy $ξ$ and Specific Angular Momentum $h$.

I had a good deal of difficulty following your work. We are likely chasing a sign error, so details matter. But what is presented here seems to be light on details.

I did this using the velocity $v = 25,000 ft/s$ and $r_{entry} = 300,000ft + 20,902,230ft = 21,202,230 ft$.

So far, so good. We have the final radius computed from Earth radius plus final altitude.

$$ξ = \frac {v^2} {2} - \frac {μ} {r}$$
Where $μ = M_⊕*G = 1.408*10^{16} ft^3/s^2$

Again, this looks good. We are computing specific energy (energy per unit mass) based on kinetic energy plus potential energy. Potential energy relative to a reference at infinity is negative, hence the minus sign.

Plugging in the values above, I got $ξ = -3.5158 * 10^8 ft/s$

Probably that is just a typo, but specific energy should be in ft/s². Let us double-check your numbers.

Sanity check: Escape energy at a particular altitude is the additive inverse of potential energy there. Google says escape velocity is about 36,000 feet per second. Escape energy = 1/2mv^2. Which yields -6.48 * 10⁸ft/s² PE. Starting velocity adds 3.125 * 10⁸ for a result of 3.36 * 10^8 ft/sec². Close enough. Sanity check passes. It is plausibly just a typo in the units.

It can't really be a velocity because nothing is moving that fast in this problem.

Then using $ξ$ I solved for the velocity $v_{instant}$:
$$v^2 = 2(ξ + \frac {μ} {r_{instant}})$$

That looks like the right equation. Let me rewrite it in a way that fits my expecations...
$$\frac{v_{initial}^2}{2} = \frac{v_{final}^2}{2} + PE_{final} - PE_{initial}$$
where $PE_{initial}$ is computed based on $r_{instant}$. You write $+ \frac {μ} {r_{instant}}$. I write $-PE_{initial}$. It's the same thing. No sign error there.

Since $r_{instant} = r_{initial} = 100$ nautical miles is greater than $r_{entry} = r_{final} = 300,000$ feet we should expect that initial PE is greater than (i.e. less negative than) final PE. Accordingly, $v_{initial}$ should be less than $v_{final}$.

Where is your calculation for $r_{instant}$?
Where is your actual computation of the result for $v_{instant}$, aka $v_{initial}$?

At this point, I am suspicious that your arithmetic did not match the formulas that you have presented here.

 

[Ray Vickson]

A space vehicle enters the sensible atmosphere of the Earth (300,000 ft) with a velocity of 25,000 ft/sec at a flight-path angle of -60 degrees. What is its velocity and flight-path angle at an altitude of 100 nautical miles during descent?

(Assuming no drag or perturbations, two body orbital mechanics)

My answers I am getting are V = 25,370.7 ft/s at a flight-path angle of -60.029 degrees.
The correct answers the book states are: V = 24,618 ft/s at a flight-path angle of 59 degrees and 58 minutes.

There is a clear correlation between my answers and the correct answers, because it seems that my answers are numerically mirrored over the 25,000 ft/s and the 60 degrees.

If someone could work this out mathematically so I can see what I am doing wrong, or just explain if I'm making a simple conceptual mistake, that would be appreciated.

Thank you in advance,
Adrian

I don't get this problem. 100 nautical miles is about 607,611.5 feet, so is waaaaay above the 300,000 foot height of the atmosphere. How can you descend from 300,000 ft. to 607,000 ft?

 

[AdrianGriff]

I don't get this problem. 100 nautical miles is about 607,611.5 feet, so is waaaaay above the 300,000 foot height of the atmosphere. How can you descend from 300,000 ft. to 607,000 ft?

It is not intending that 100 nmi is lower than the entry altitude, it is asking what the velocity and flight-path angle were previously when higher in orbit assuming it is on an entry path back to Earth.

 

[AdrianGriff]

I had a good deal of difficulty following your work. We are likely chasing a sign error, so details matter. But what is presented here seems to be light on details.

So far, so good. We have the final radius computed from Earth radius plus final altitude.Again, this looks good. We are computing specific energy (energy per unit mass) based on kinetic energy plus potential energy. Potential energy relative to a reference at infinity is negative, hence the minus sign.

Probably that is just a typo, but specific energy should be in ft/s². Let us double-check your numbers.

Sanity check: Escape energy at a particular altitude is the additive inverse of potential energy there. Google says escape velocity is about 36,000 feet per second. Escape energy = 1/2mv^2. Which yields -6.48 * 10⁸ft/s² PE. Starting velocity adds 3.125 * 10⁸ for a result of 3.36 * 10^8 ft/sec². Close enough. Sanity check passes. It is plausibly just a typo in the units.

It can't really be a velocity because nothing is moving that fast in this problem.

That looks like the right equation. Let me rewrite it in a way that fits my expecations...
$$\frac{v_{initial}^2}{2} = \frac{v_{final}^2}{2} + PE_{final} - PE_{initial}$$
where $PE_{initial}$ is computed based on $r_{instant}$. You write $+ \frac {μ} {r_{instant}}$. I write $-PE_{initial}$. It's the same thing. No sign error there.

Since $r_{instant} = r_{initial} = 100$ nautical miles is greater than $r_{entry} = r_{final} = 300,000$ feet we should expect that initial PE is greater than (i.e. less negative than) final PE. Accordingly, $v_{initial}$ should be less than $v_{final}$.

Where is your calculation for $r_{instant}$?
Where is your actual computation of the result for $v_{instant}$, aka $v_{initial}$?

At this point, I am suspicious that your arithmetic did not match the formulas that you have presented here.

I noticed that your units for $ξ$ were not $ft^2/s^2$ are they not supposed to be that way? From the equation for $ξ$ I see no reason why both units distance and time should not be squared.

Also, when you did the 'sanity check' you said Escape energy = 1/2mv^2 and that you got an energy of -6.48 *10^8 ft/s^2 and I am wondering how you got the value from this when we do not know the mass of the object in orbit. Also where was the initial energy from the sanity check that you added from? Is that a constant you had used, or was it calculated from something else?

 

[Ray Vickson]

It is not intending that 100 nmi is lower than the entry altitude, it is asking what the velocity and flight-path angle were previously when higher in orbit assuming it is on an entry path back to Earth.

OK, that makes sense, especially if you are supposed to neglect air friction.

 

[AdrianGriff]

Did you start by standardizing your units of measure? I notice that in the original condition of the spacecraft , the altitude is given in ft, but the question asks for speed and angle at an altitude given in Nautical Miles. What did you get for that? (Just want to make sure you didn’t pull a Mars Climate Orbiter.)

Yes I standardized all units haha

 

[jbriggs444]

I noticed that your units for $ξ$ were not $ft^2/s^2$ are they not supposed to be that way? From the equation for $ξ$ I see no reason why both units distance and time should not be squared.

You are correct, of course.

Also, when you did the 'sanity check' you said Escape energy = 1/2mv^2 and that you got an energy of -6.48 *10^8 ft/s^2 and I am wondering how you got the value from this when we do not know the mass of the object in orbit.

Specific energy -- I assumed 1 kg mass (or one pound, or one slug or one whatever unit mass).

Also where was the initial energy from the sanity check that you added from? Is that a constant you had used, or was it calculated from something else?

Let me run through that calculation for you. Please try to emulate this. When we ask you to show your work, please show your work.

We want the total energy of something moving at 25,000 feet per second at a near-earth altitude. The strategy is to compute KE + PE = total energy. We want "specific energy", i.e. energy per unit mass. So we will assume a unit mass.

KE is given by $\frac{1}{2}mv^2$. So we compute $\frac{1}{2}* 25,000^2$ = $\frac{1}{2} * 625,000,000$ = 312,500,000 ft²/s².

PE is well approximated by escape energy. Escape energy is $\frac{1}{2}mv^2$ where the v is low Earth escape velocity, approximately 36,000 ft/sec. So we compute $\frac{1}{2}*36,000^2$ = $\frac{1}{2} * 1,296,000,000$ = 648,000,000 ft²/s². But PE taken against a reference at infinity is negative, so negate that.

KE + PE = 312,500,000 - 648,000,000 = 335,500,000 ft²/s²

So... What number did you get for $r_{instant}$.