The picture of this prob here was lost but you can find it below ,sry I was wondering how to exactly calculate this one ... Ok, let's presume some dug a hole thru our mama earth. Ok, YOU jump into hole (Let's count out the airs friction ,earths turning and movement) 1. How long is it gonna take until you're on the other side ? 2.What's the speed you reach at the center point of earth ? The problem for me was to derive the G in each point since all the mass what's 'above' (or below ,respectively) and also around you differs the G you are 'experiencing' .. I know the're are many good mathemathics out there since discussion of 'hydrogencars' ,so PS. pls jump head-first to get out head up PPS. Earths mass is 5.972e24 kg
Ad1 - You won't get to the other side cause you'll burn your guts in the center Ad2 - Dunno but quite fast Btw i think in center (if you won't die due the temperature , you'll won't pass the center due to the gravity at the other side ... you will just spin or stay still in the center of the earth )
Yes... Vojtas - you're right! you won't fly to the otherside ("how long, how loooong? ") , you'll fly cross the center, and slow down, stop, and return to center
that is if there's resistance on the object if we think about an ideal resistance less environment, and about a place where the distance between the centre and the two entrances is exactly the same... in such a case the object would be slung far enough back up to be able to say the energy states at P1 and P2 are the same btw, IS there a centre when you cut a hole through it? j/k greetze, Zembla
vojtas : OMG man!! the freaking tube is so damn thick there's no fking way you even feel the mega-hot liquid metal around you as you go go gooo!! ...so this situation happens is in 'ideal' conditions AND on an IN (hehe) ideal ball (which this is?!? )so to say ... beryl : Negative.Your potential energy transfers to kinetic and when you pass the midlle point ,then your kinetic energy equals the potential earth causes ,and thus ,you arrive to other side ... zembla : CC buddy ,like there has been a virus in the center of my nostral whole my life trying to get to the edges
"Look mommy, I created a "Perpetuum Mobile"!!!!" You'd reach very high speeds, but due to air-friction you'd loose a lot of E. Think of it this way: If you fly at 10.000 meters, shut down the engine, push down the stick, dive to 8.000 meters and then pull up, would you reach 10.000 meters without starting your engine? And then, even if this was done in a vacuum, you'd be pulled apart due to the diverging gravitational pulls.
1. 1/2 time of sattelite's full turn time at h~=0km (disregarding Earth's atmosphere), that is 40+ min 2. sqrt(g*R) ~= 7.9 km/s R=Earth's radius, ~6400 km The problem is much simpler that you think. The 'above' mass does not affect g. The basic equation is div(g) = q*density, q - normalizing coefficient, and under conditions of spherical symmetry, you would finally get g(r) = g*(r/R)^2
zembla, there's no need to punch the centre. in general the problem is to calculate G within earth body. that's a problem for non-math guy like me. i'd go AROUND this problem. what is period of earth satellite with perigei = apogee (circle orbit) and orbit radius = planet radius? well, that sounds more optimistic. [edited] first cosmic speed (speed to stay on orbit) = square root from radius multiplied by G on the surface. r = d / 2 = 12754km / 2 = 6377km = 6.377*10e6m sqrt(6.377*10e6m*9.81m/(s*s)) = 25011.671275626505m/s = 90042.016592255418km/h perimeter (length of the circle) is pi*d = 3.14159265358979324*12754km = 40067.872703884223km (wow, it's a length of equator familiar to me!) so, the time the satellite make a circle is 40067.872703884223km / 90042.016592255418km/h = 0.44499084116837173h = 26'41.967" so the half-period is 13'20.983" and the speed at the centre is first cosmic speed. i must have made a million of errors (and lazy to correct them), but i hope you got my logics in general.
dankes, you are my personal enemy! i typed it for about 5 minutes for nothing, because you are first who posted the right calculations
Theres no power without direction. So what direction is in center? All or nothing? BTW U all should read Stephen Hawkins or Douglas Adams but nothing between it!
Don't read this siht daer t'no on't read thi As you approach the center of Earth during your "downward leg", is not it true that the majority of mass is above you, therefore, you are no longer tugged downward at 1 gee? I dunno if I said that correctly. What is the gravitational force at Center Of Mass? Beats the hell out of me.
gravitation at the centre of mass is zero. in partial situation of hole sphere, gravitation is absent within the sphere at all.
Re: Don't read this siht daer t'no on't read thi Hmmmm yeh as far as I remember it: gravitational force is centralised in the core of the object when you're trying to measure the gravity of an object that is placed outside the perifery of the object with the gravity field etc... dunno about the inside of it though, think it remains the same in general... a bit like with the Coulomb forces greetze, Zembla
wtg exec- & dankes!! I salute you ,still ....I want the top speed at center , and the time to get to the "otheeer sideee" @biles : Of course the gravitation is zero at the middle cause all particles around you have on opposite one which 'defeat' eachother .. -1 + 1 = 0 aagh PPS. I suggetst for you not only to jump in to hole head first ,but also jump from a 2 meters high stand to get few meters up on the otherside for someone to grab you before you sunk in again
