# Escape Velocity and Centripetal/Centrifugal Acceleration

This is just a reality that I have stumbled upon that I'm sure was well-known, but I still found it interesting. I apologize if this is second-nature to physics experts.

I was responding to a post on a different thread that claimed you could make a tunnel around the Earth and if you sent a train or something around that tunnel at 11.2 km/s, the people inside would feel weightless. Of course, this isn't the case; they only need to be traveling at what the orbital velocity would be at the surface of the Earth. I corrected him, and also wanted to be able to tell him what would happen if the train WERE going 11.2 km/s, and I calculated it, and as it turns out, the people on that train would experience 1G, but toward the top of the train. I though maybe this was a coincidence, so I actually did the math and it turns out that it is EXACTLY 1G. Like I said, I'm sure this is well-known, but I though it was a cool fact. At escape velocity, you would experience exactly 1G toward the roof. It would feel as though you were still on Earth, but the ceiling of the train would become the floor. I further delved into the equations and came up with why this was the case.

The acceleration due to gravity is GM/r². Centripetal acceleration can be calculated as v²/r. Escape velocity is √(2GM/r), and if we plug in the escape velocity into the centripetal acceleration equation, we get 2GM/r². I found that to be very cool. Acceleration due to gravity is GM/r² while centripetal acceleration with a radial velocity equal to escape velocity would be 2GM/r². Gotta love physics.

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• Dale

It would feel as though you were still on Earth, but the ceiling of the train would become the floor. I further delved into the equations and came up with why this was the case.

somehow i could not feel the thrill of 'new physics'....or i missed it....normally if you rotate anything on earth say a man sitting inside the rotating body will need a force to keep it on that path and the reaction of the top of the container will provide it , otherwise he will fly off tangentially. so whatever be the magnitude of the velocity ...the force needed will be mass times v^2 divided by r. thanks and pl explain what was amiss.

Dale
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I actually did the math and it turns out that it is EXACTLY 1G. Like I said, I'm sure this is well-known, but I though it was a cool fact.
I didn’t know it! Definitely interesting

CWatters
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Thought I would do the calculation for the moon and it appears to give the same answer....

The escape velocity of the moon is 2.38 km/s. The radius of the moon is 1737km.

a = v2/r = 23802 / 1737,000
= 3.26 m/s2

The actual acceleration due to gravity is 1.63 m/s2 so the net upward acceleration is 3.26 - 1.63 = 1.63m/s2 or 1 moon g upwards.

• lekh2003
Yes, it would work anywhere. Basically, if you were in some vehicle that was traveling in a circle with a radius equal to the distance from the center of any given mass at a radial velocity equal to the escape velocity from that mass at that height, the net acceleration would be equal to the gravitational acceleration - just in the opposite direction since the resulting radial acceleration is exactly equal to twice the gravitational acceleration from that same height.

jbriggs444
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One could derive it from the related fact that orbital energy for a circular orbit is half of escape energy. Double the energy and you've multiplied speed by ##\sqrt{2}##. Multiply speed by ##\sqrt{2}## and you've doubled centripetal acceleration.

It looks like a special case of the virial theorem.

sophiecentaur