Suppose a spacecraft is returning from the Moon but screws up it’s course correction so that instead of grazing the atmosphere obliquely, it slams into the atmosphere dead on. What G-force will the capsule experience?
Doesn’t it depend totally on what angle it’s going to hit the Earth, and how fast it’s going?
You mean coming in for a landing straight down, drilling into the Earth, so to speak? Doesn’t the G-forces it experiences depend totally on how fast it’s going, then? Do you want to specify a speed, or should we just assume normal re-entry speed, whatever that may be?
Also, there’s the problem that if it’s going too fast, it’ll burn up. Browsing around on Google, it looks like a spacecraft would have to worry about burning up before it would have to worry about maximum tolerable G-forces.
So do you want to stipulate that the spacecraft is flame-proof?
Remembering from the film Apollo13, a re-entry in excess of 10 Gs will result in burn-up of the capsule.
It wouldn’t be the acceleration (‘g forces’) that would result in the burn up, but the velocity that would be required to get that acceleration. The speed at which it would burn up would also obviously be determined by the material that is the heat sheilds.
Lumpy, your question can’t be answered without first knowing what the ship looks like. Think of it as hypersonic flight; the forces you’re looking for would be manifested as drag. The ship is ‘flying’ from a vacuum into the atmosphere, and as the atmospheric density and pressure changes, so does friction. A SR-71 or Concorde shaped space ship would be better off than an equally big ship shaped like a bell.
Depends on velocity. The reason (I imagine) that they want to hit the Earth at a sharp angle is so that they are slowly immersed in a denser and denser atmosphere as they get closer and closer (there’s no solid boundary for the atmosphere). The heavily diluted atmosphere at 250 kilometers up (or whatever) will start slowing the capsule down. That way, by the time they’re at 200 km, they’re travelling significantly slower than they initially were… and so on, all the way down.
continuing from what wikkit said…
You have yourself a real Catch-22 here, son. In non-physics terms, G-forces result when you suddenly increase speed, or (in the case of reentry) suddenly slow down. This can be demonstrated in any car ride.
Now, if you’re in something shaped like a Concorde, as wikkit suggested, your g-forces will be on the mild side, because you’ll sail straight through without any serious slowing down. Which means you’ll crash pretty bad at the end of the ride.
In something more bell-shaped, like the Apollo, there’s some serious slowing down, which increases both the g-forces and the friction (friction= chances of burning up).
The trick is to balance all these things so that you slow down at a reasonably steady pace to minimize the g-forces and the chances of crashing or burning.
Ok, some parameters to go by:
- Re-entry speed is 7 miles per second, the freefall speed attained by something returning from a translunar flight path, like the Apollo capsules.
- Ballistic reentry, also like the Apollo capsules.
- Let’s assume a robustly shielded capsule capable of taking whatever heat is generated.
As a rough and ready calculation, I started with an initial speed of 7 miles per second, a nomimal reentry at 90 miles altitude, and sub-sonic terminal velocity reached by 10 miles altitude. To slow down from 7 miles per second over 80 miles, my math gives me a decelleration required of 45 Gs (!) The problem with that figure is twofold: First, you would not decellerate evenly the whole time, the peak Gs could well be higher. Secondly, I’m not sure the atmosphere could exert enough drag on the capsule to slow it down that fast; it might still be doing a couple of miles per second when it hit the ground. It’s the problem with real-life aerodynamics that I don’t know how to calculate.
The biggest problem you have Lumpy is that you aren’t traveling straight down from 90 miles to 10 miles, you are descending at a shallow angle. So, it might take you 1000 miles to decelerate to your parachute deployment speed.
douglips: the whole point of this discusion is what would happen if you hit it dead on, rather than a shallow angle.
Lumpy: I haven’t done your math, but that doesn’t sound unreasonable. I guess the final answer is this: What happens if an adequately shielded, massive object strikes the earth’s atmosphere dead on? A meteor crater.
Wikkit:
Nit note:
You can’t have a meteor crater, since meteors don’t hit the earth, just burn up when they hit the atmosphere. MeteorITEs hit the earth. Even big, dangerous asteroids are meteorites if they collide with lttle ol’ us.
Here’s a back-of-the-envelope calculation (sorry for the lack of pretty Greek letters):
First, the atmospheric pressure p varies approximately exponentially with altitude z: p ~ p[sub]0[/sub] e[sup]-z/h[/sup], where h is the scale height of the atmosphere, h ~ 7500 m. The temperature varies with altitude, but not by anything like an exponential, so I’ll pretend it’s isothermal. Then the density x (pretend that’s a rho) also varies exponentially with altitude,
x ~ x[sub]0[/sub] e[sup]-z/h[/sup].
The force of air resistance on an object is a complicated function of the object’s shape and size and the air density. I think that for large objects moving at high speed, the drag force can be approximated as
F ~ 0.5 x A v[sup]2[/sup],
where A is the object’s cross-sectional area, x is the air density, and v is the object’s speed. (When the object moves a distance d, it does work F d ~ 0.5 x A d v[sup]2[/sup]. The right-hand side is the kinetic energy of a cylinder of air of volume A d moving at speed v.)
So, ignoring gravity (adding gravity is an Exercise for the Student :)), we have the (order-of-magnitude) differential equation for the deceleration of the spacecraft
ma = F = 0.5 x A v[sup]2[/sup] = 0.5 x[sub]0[/sub] A v[sup]2[/sup] e[sup]-z/h[/sup].
Now if the spacecraft is entering the atmosphere at the oblique angle q (the angle from vertical; pretend that’s a theta) then dz/dt = v cos q. So v = (dz/dt) sec q and, assuming q doesn’t change rapidly with time, a = (d[sup]2[/sup]z/dt[sup]2[/sup]) sec q.
This is enough to estimate answers to the questions posed. For example, you can estimate a as a function of v, z, and q. To solve the later question (what should the spacecraft trajectory look like for constant deceleration), let a be the desired deceleration and solve the equation to give v as a function of z (and the constant parameters a, m, A, h, and x[sub]0[/sub]).
Now the spacecraft trajectory (q as a function of z) must be chosen to match this velocity curve. That is, the spacecraft must spend the right amount of time between altitude z+dz and z to decelerate from v(z+dz) to v(z): a = dv/dt = (dv/dz)(dz/dt) = (dv/dz) v cos q. So cos q = a / (v dv/dz). Plugging in v(z) gives q as a function of z,
cos q = (2 h A x[sub]0[/sub] / m) e[sup]-z/h[/sup].
For the OP, q = 0, so dz/dt = v and dv/dt = a. The differential equation can be integrated once to give
v = v[sub]0[/sub] exp[-(x[sub]0[/sub] A h / (4 m)) e[sup]-z/h[/sup]].
Plugging v into the differential equation gives a as a function of z:
a = x[sub]0[/sub] A v[sub]0[/sub][sup]2[/sup] / (4 m) exp[-x[sub]0[/sub] A h / (2 m) e[sup]-z/h[/sup]] e[sup]-z/h[/sup].
This is maximized at z[sub]max[/sub] = h ln (x[sub]0[/sub] A h / (2 m)), with a[sub]max[/sub] = v[sub]0[/sub][sup]2[/sup] / (2 h e).
Plugging in some real numbers – x[sub]0[/sub] ~ 1 kg/m[sup]3[/sup], v[sub]0[/sub] ~ 10000 m/s, h ~ 7500 m, m ~ 5000 kg, A ~ 10 m[sup]2[/sup] (I think these last two are about right for Apollo) – and plotting a(z) gives a peak acceleration of about 250g at about 15km altitude.
[sup][sub]OK, so it was a large envelope.[/sub][/sup]
Google search, meteor crater: about 33000 results. Same for meteorite crater, 20600.
I know the difference between a meteor, meteorite, meteoroid, etc. They are commonly called meteor craters, even if they aren’t.
I bet you’ve never drank out of a styrofoam cup.