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Old April 26 2010, 12:28 PM   #1
SilentP
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Space Jumping?

Essentially can it be done like we see in Star Trek XI? Does the small size of the person not create enough drag to start burning up regardless of the angle of entry, or does the atmosphere indeed burn up anything dropping straight towards the size?
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Old April 26 2010, 01:32 PM   #2
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Re: Space Jumping?

In a word, yes, but not from an orbital velocity. The reason things burn up on entry into the atmosphere has everything to do with the object's velocity relative to the atmosphere, and not with the fact that you're going from space into an atmosphere. For example, the space shuttle is trying to slow down from about 17,000 mph (orbital velocity), and that friction against the atmosphere is what creates all the heat. In contrast, Spaceship One also went into space, but it basically went straight up and down, and was not going anywhere near as fast, so it didn't need any heat protection.

In ST XI, since they were dropping down on the drill, which was stationary over the Earth, they wouldn't be going at orbital velocities, so it should be possible.
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Old April 26 2010, 01:45 PM   #3
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Re: Space Jumping?

Ah, so the velocity is what matters most here then.

Is it velocity towards the surface that matters most, or relative to the surface (i.e. 'groundspeed') or a combination of both, since it's a case of how fast an object tries to move through the atmosphere?
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Old April 26 2010, 01:58 PM   #4
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Re: Space Jumping?

The surface really doesn't matter, it's your speed relative to the atmosphere that counts.
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Old April 26 2010, 02:50 PM   #5
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Re: Space Jumping?

SilentP wrote: View Post
Essentially can it be done like we see in Star Trek XI? Does the small size of the person not create enough drag to start burning up regardless of the angle of entry?
In the case of motion much slower than the speed of sound, we have Stokes Law, which says that the magnitude of the drag force is directly proportional to both the speed of the object and the diameter of the object.

That force causes deceleration according to Newton's law, a = F/m

... and m is the mass, no doubt proportional to the cube of the diameter if construction materials retain the same density.

So all in all, that is deceleration = (constant * speed * diameter ) / diameter ^3

= constant * speed / diameter^2
= constant * v/d^2

The smaller an object is, the stronger it will be decelerated as it moves through the atmosphere.

Compare throwing a grain of sugar with throwing a bag of sugar. One is just a 100 times the diameter of the other. But you'll be lucky to throw the 1mm diameter grain across the room to hit the opposite wall because it will be decelerated much stronger than the 100mm diameter bag.



Now, that deceleration is a change of speed, which changes the kinetic energy of the object. This energy is converted mostly into heat over the surface and in the passing air.

KE = 1/2 m.v^2

rate of heating = d(KE)/dt = m.v.a =( constant * d^3) * v * (constant * v / d^2)

= constant * v^2 * d

The surface area is proportional to d^2, so the heat generated per unit area = heat / d^2
= constant * v^2 / d

So the smaller an object is, the quicker it's surface temperature will rise as it is moving. So the more likely the surface will burn.

Of course, this applies to slow movement of spherical objects, but what happens over mach one with uneven shaped bodies it is not too dissimilar; it just changes the constant.
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Last edited by Jadzia; April 26 2010 at 03:12 PM.
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Old April 26 2010, 08:43 PM   #6
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Re: Space Jumping?

Jadzia wrote: View Post
SilentP wrote: View Post
Essentially can it be done like we see in Star Trek XI? Does the small size of the person not create enough drag to start burning up regardless of the angle of entry?
In the case of motion much slower than the speed of sound, we have Stokes Law, which says that the magnitude of the drag force is directly proportional to both the speed of the object and the diameter of the object.

That force causes deceleration according to Newton's law, a = F/m

... and m is the mass, no doubt proportional to the cube of the diameter if construction materials retain the same density.

So all in all, that is deceleration = (constant * speed * diameter ) / diameter ^3

= constant * speed / diameter^2
= constant * v/d^2

The smaller an object is, the stronger it will be decelerated as it moves through the atmosphere.

Compare throwing a grain of sugar with throwing a bag of sugar. One is just a 100 times the diameter of the other. But you'll be lucky to throw the 1mm diameter grain across the room to hit the opposite wall because it will be decelerated much stronger than the 100mm diameter bag.



Now, that deceleration is a change of speed, which changes the kinetic energy of the object. This energy is converted mostly into heat over the surface and in the passing air.

KE = 1/2 m.v^2

rate of heating = d(KE)/dt = m.v.a =( constant * d^3) * v * (constant * v / d^2)

= constant * v^2 * d

The surface area is proportional to d^2, so the heat generated per unit area = heat / d^2
= constant * v^2 / d

So the smaller an object is, the quicker it's surface temperature will rise as it is moving. So the more likely the surface will burn.

Of course, this applies to slow movement of spherical objects, but what happens over mach one with uneven shaped bodies it is not too dissimilar; it just changes the constant.
Awesome!
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