# Thread: As usual - can't do the maths

1. Established Member
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## As usual - can't do the maths

Hi,

Can anyone tell me how fast I'll be going when I hit the ground at sea level if I fall from 100 km up through the Earth's gravity field.

Oh yeah -and assume there's no atmospheric drag (i.e. vacuum) please.

I understand it's a bit tricky as the 9.8 m/sec force of gravity at sea level decreases as you get higher from the Earth's centre of mass.

2. From Wikipedia, Terminal velocity.
Based on wind resistance, for example, the terminal velocity of a skydiver in a free-fall position with a semi-closed parachute is about 195 km/h (120 mph or 55 m/s). This velocity is the asymptotic limiting value of the acceleration process, since the effective forces on the body more and more closely balance each other as the terminal velocity is approached. In this example, a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on. Higher speeds can be attained if the skydiver pulls in his limbs (see also freeflying). In this case, the terminal velocity increases to about 320 km/h (200 mph or 90 m/s), which is also the terminal velocity of the peregrine falcon diving down on its prey.
I Don't know if this helps you at all.

3. ## beddie bye

Iam going to bed, so i will look later. But on the one hand i would cheat and use 9.8m.s^-2 anyway. You probably want to integrate the path F=GMm/r^2 (r1=100x10^3 to r2=0) this i think will give you your kinetic energy just before hitting ground.

4. ## Use gravitational potential energy

Originally Posted by Cheap Astronomy
Hi,

Can anyone tell me how fast I'll be going when I hit the ground at sea level if I fall from 100 km up through the Earth's gravity field.

Oh yeah -and assume there's no atmospheric drag (i.e. vacuum) please.

I understand it's a bit tricky as the 9.8 m/sec force of gravity at sea level decreases as you get higher from the Earth's centre of mass.

Look in your textbook for an expression for the gravitational potential energy of an object of mass "m" which is located a distance "R" from another object of mass "M". Compute the gravitational potential energy of a human when it is (100 km + R_earth) away from the center of the Earth, and when it is (R_earth) away from the center of the Earth. The difference in GPE will be equal to the kinetic energy of the person just before it strikes the ground.

5. kzb
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distance = 0.5at^2

and

velocity = at

100,000m = 0.5*9.81ms^-2*t^2

t^2 = 100,000/(0.5*9.81)= 20,387

t= SQRT(20,387) = 142.8 seconds

v= 9.81 * 142.8 = 1,401 metre per second

That's about 3,134 mph. I've neglected the change in acceleration due to gravity with height, as it's quite small over 100km.

6. Order of Kilopi
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The acceleration due to gravity drops from 9.8 m/s^2 at sea level
to 9.5 m/s^2 at 100 km. That isn't much, but integrated over the
time of the fall, it should make a significant difference in the speed.

-- Jeff, in Minneapolis

7. From 100 km altitude you achieve great speed prior to reaching thicker atmospheric layers that will slow you to terminal speed.

8. Order of Kilopi
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Originally Posted by StupendousMan
Look in your textbook for an expression for the gravitational potential energy of an object of mass "m" which is located a distance "R" from another object of mass "M". Compute the gravitational potential energy of a human when it is (100 km + R_earth) away from the center of the Earth, and when it is (R_earth) away from the center of the Earth. The difference in GPE will be equal to the kinetic energy of the person just before it strikes the ground.
Stu. Yep. That works for photons, too...letting hv=mc2. find the "mass", add the extra kinetic energy to the photon at the surface, and it pops out the gravitational blue shift.

9. Originally Posted by kzb
100,000m = 0.5*9.81ms^-2*t^2

t^2 = 100,000/(0.5*9.81)= 20,387

t= SQRT(20,387) = 142.8 seconds

v= 9.81 * 142.8 = 1,401 metre per second
Originally Posted by Jeff Root
The acceleration due to gravity drops from 9.8 m/s^2 at sea level
to 9.5 m/s^2 at 100 km. That isn't much, but integrated over the
time of the fall, it should make a significant difference in the speed.
The changing acceleration makes only a minor difference. Instead of taking 142.8 s and going 1401 m/s, the numbers are 144.7 s and 1390 m/s.

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Thanks all.

Using reverse logic it seems then that if I get shot out of a cannon at 1.4km/sec I will make about 100km altitude before falling back (once again assuming vacuum conditions).

And on that basis can you tell me what altitude I get to if I launch at 11.1999km/sec (i.e. just a bit less than escape velocity)?

Thanks heaps (and hope I'm not breaking some 'one question per thread' rule here)

11. Order of Kilopi
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judgement about when it is best to reuse an existing thread and when
it is best to start a new one.

However, using a humongous number of significant digits in a value that
is a premise to a question can get you locked in the pillory all day.

As a practical response, look up "Hill sphere".

-- Jeff, in Minneapolis

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## Potential energy

Look at it this way: kinetic energy is proportional to the square of speed. Potential energy is proportional to the inverse of the distance.

Letīs round the numbers a bit. As Earthīs escape speed is 11,2 km/s, whose square is 125,44 km2/s2, round the latter to 125 (having escape speed of roughly 11,18 km/s). And round the 6378 km radius to 6400 km. Then the product is 800 000 km3/s2.

Shoot something up at 11,0 km/s. Square is 121 km/s. The difference from escape speed squared is 4 km2/s2.

Therefore an object shot up at 180 m/s below escape speed rises to 200 000 km distance from centre of Earth.

If the speed were 18 m/s less than escape speed, it is easy to see that the square would be about 124,6 km2/s2, and the height would be 2 000 000 km. But then you also have problems with Hill distance or, in other words, perturbations from Sun.

13. kzb
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<<The changing acceleration makes only a minor difference. Instead of taking 142.8 s and going 1401 m/s, the numbers are 144.7 s and 1390 m/s.>>

Damn. If I'd stuck to 9.8 instead of 9.81 and reported everything to 2 significant figures as I should, I'd have been right. I'll learn a lesson there.

14. The equation for falling from any distance without resistance in Newtonian gravity is v^2 = 2 G M (1 / r - 1 / s). 's' is the distance one falls from and 'r' is the distance one falls to, the radius of Earth in this case. If we rearrange, we can get v^2 = 2 G ME (s / rE s - rE / rE s) = 2 G ME (s - rE) / rE s = 2 (G ME / rE^2) (rE (s - rE) / s) = 2 gE (rE (s - rE) / s), where s - rE = 100 km for what you're asking and s = rE + 100 km = 6371 km + 100 km = 6471 km. So we have v^2 = 2 (9.81 m / sec^2) [(6.371*10^6 m) (100000 m) / (6.471*10^6 m)], giving v = 1390 m / sec = 1.39 km / sec.

The other is gained using v^2 = 2 g (s - r) = 2 (9.81 m / sec^2) (100000 m) = 1401 m / sec for constant acceleration, which is close enough for falling small distances compared to the radius of the Earth, where r approximates s and falls out of the first equation.

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Woo hoo - Hill's Sphere - awesome. Thanks all. This forum kicks a\$#

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