If I dug a tunnel thru the earth and jumped in how fast would I be moving when I reach the center of the earth

Question

You can assume no wind resistance. Do account for changes in gravity. Show your work! And no this is not for homework!

Answer 1

The practical solution is that you would not make the trip to the center, you would be burnt to ashes on your way down.

The theoretical solution (with all of the assumptions) is that you would reach zero velocity at the very center, but you might need to oscillate through the center several times to do this. The math involved is beyond the format of this site.

Answer 2

I don't have time to show all the equations but I can tell you this at the exact center of the earth gravity from the earth would be pulling (nearly) equally on you in all directions. Because of this you would be moving at a velocity very close to 0. When you first jumped you would actually oscilate a few times up and down before you finally reached a velocity of near 0. also because the earth is of course not a perfect sphere you would never reach a total stop.

Answer 3

The answer is that your final velocity -- V(final) -- would be 1,180.85 m/s (see below for explaination for how you derive that answer)

The Earth's equatorial radius (or semi-major axis) is the distance from its center to the equator and equals 6,378.135 km (or 6,378,135 m), so that is the distance that you would need to travel. We will call that "d," for distance.

V(final)^2 = V(initial)^2 + 2ad
(a = gravity = -9.80 m/s^2)
V(initial) = 0 m/s because you're starting from rest.

Putting in the numbers you get:

V(final)^2 = (0 m/s)^2 + 2(-9.80 m/s^2)(6,378,135 m)

V(final)^2 = -125,011,446 m^2/s^2

You now need to take the square root of -125,011,446 m^2/s^2 to solve for V(final).

V(final) = 11,180.85 m/s

I know that that number seems large, but you'd be falling for a long, long, long time, and the acceleration is related to time.

And because you are assuming no air resistence, there is no terminal velocity calculation needed. The terminal velocity of an object falling towards the ground, in non-vacuum, is the speed at which the gravitational force pulling it downwards is equal and opposite to the atmospheric drag (also called air resistance) pushing it upwards. Because there is no air resistence in your problem, there is no atmospheric drag, and therefore no terminal velocity. As such, the object keeps accelerating at 9.8 m/s^2.

Answer 4

32ft p/sec.(distance) - And a serious person would have to account for changes in gravity. - The closer you get to the source, the stronger the pull. - Until "critical mass" is achieved. - I think it remains at 86,000 miles p/sec. - A fraction of the distance is a fraction of the speed. I don't have a calculator - but neither do you. - The center of the Earth is approx. 6,000 miles beneath us. - A better question would be how long would it take to get there.

Answer 5

if u do away with air resistance, i think of you will end on the different area of the earth at an analogous top at which you jumped relative to the middle of the earth... that's conservation of potential... u initiate off w/ specific means potential, then will become kinetic... and returned to means, so rather you will oscillate from area to area, like a spring

Answer 6

You wont reach there.
You ll be baked, roasted and evaporated much before that. Heat inside is just too intense- in excess of 2000 degC

Answer 7

You would be going at terminal velocity which I believe is approximately 180mph.....

Do I win a prize?

Answer 8

3.9 meters per second square

Answer 9

hello u'l be dead before u know!