how fast must i run to jump off the planet and use only the earths gravity to continue "flying" around the planet (assuming that mountains don't get in the way and that there is no wind)?
About 18,000 MPH would do it for that's what it takes to reach orbit with any rocket that is sent into space. Better be in damn good shape to reach that escape velocity! Please Register or Log in to view the hidden image!
what do you mean by "flying around the planet? you want to be flying above a statis point on the planet (geostationary satellite) or do you want to see the land scrolling past underneath you? 18,000mph would have to be the vertical component needed to escape.. there'd also be a horizontal (relative to the earth's surface) component if you want to scroll the land..pythagoras should give that to you
You have a units conversion mistake, James... try about 28000 km/hr. +/- up to 1700km/hr, depending on direction and latitude (unless assuming a nonrotating Earth)
Oops. You're right. The 8000 m/s is right. But then I divided by 3.6 to convert to km/hr instead of multiplying. 28000 km/hr is about right.
I guess you would burst into flames prior to achieving the desired speed though (glad Leblanc didn't hear that!).Please Register or Log in to view the hidden image!
True, but they are specially designed for such, don't achieve those speeds until they reach the upper atmosphere, etc. I'm sure you're aware that rocket capsules in orbit that re-enter the atmosphere experience tremendous heating in the process, to the point that they will completely disintegrate unless they are specially shielded with custom heat-insulating materials. I would fully expect that a human being running at sea level would indeed burst into flames long before reaching 28 Mm/hr.
Another way to say it is that you need to run fast enough so that when you jump you will fall according to the earths curvature. You have a downward acceleration of 9.81 m/s, then we need to know how many metres per second you need to travel to make 9.81 m/s represent earths curvature. I don't know the calculation needed for this though, but that would be my reasoning if I were to understand how to arrive at the results. One reasoning might be the formula for a circle, if given a inward motion, how much would I need to compensate by forward speed?
This is how I calculated it. For a circular orbit of radius \(r\) under the force of gravity the net force exerted on a mass \(m\) by the Earth (mass \(M\)) is: \(F=\frac{GMm}{r^2}=\frac{mv^2}{r}\) Re-arranging gives: \(v=\sqrt{\frac{GM}{r}}\) This is the required orbital speed. The relevant numbers at the Earth's surface are: \(G=6.67\times 10^{-11}, M=6\times 10^{24}, r=6.37\times 10^6\) to get a speed in metres per second.
Thanks for the equations! I was probably a bit off with my reasoning so it's nice to see what was behind the numbers.
