No.
You keep your current velocity.
If gravity went away, you would slowly drift sideways as the earth turned beneath you.
Actually, yes. Your ball and string analogy works. The other consideration is where you are on the globe when gravity is switched off. You will fly off the eath on a path perpendicular to the axis of rotation.
At the equator your rotational velocity is a bit over 1000 mph. You would travel straight up at some speed. High school physics is too far in the past for me to remember exactly how fast you would hit the ceiling, but I don’t believe you would enjoy it much. OTOH, in all likelyhood the building you are in would likely be departing with you.
As you travel further north or south you get closer to the axis of rotation and your rate of travel decreases as a function of the cosine of your latitude. At 60 degrees latitude your speed will only be about 520 mph and you would depart at a 30 degree angle relative to the surface of the Earth.
More speeds here.
Although I’d also venture that unless the same magic that switched off gravity took over holding the Earth together that the Earth would immediately become a rapidly expanding disc of debris.
Wow! Thanks all
Sorry so long getting back. Work doesn’t allow much downtime…
Great answers all, though I must say my head is spinning trying to follow them. I’m afraid I don’t have a spring scale available to experiment with, but
Hampshire got closest to the correct scenario here:
…as I have my left foot on the left peg of the bike, and although I kick with my right foot it’s to lift my body upward rather than to exert downward force on the peg, much as you would to mount a horse (or step up on a chair).
And just for the record, I wasn’t concerned about breaking the peg, just about making it too hard for the SO to keep the bike upright with that much force to one side.
I really appreciate everyone’s input. It sure provides food for thought, but confirms that the pressure sideways shouldn’t be THAT great…
-jean
The integral that Santo Rugger speaks of is impulse, the integral of force with respect to time, which is equal to the change in momentum:
[indent]I = F·Δt = m·Δv = Δp[/indent]
Where F is force, Δt is the time interval, m is mass, Δv is the change in velocity, and Δp is the change in momentum. If we divide both sides by the Δt and rearrange things a bit, we get
[indent]F = I/Δt = Δp/Δt = m·Δv/Δt = m·a[/indent]
So the amount of force exerted depends not only on the mass, but how quickly the change in momentum is applied; the quicker the change, the larger the force, even for the same amount of total energy or resultant velocity. Going back to the example of the bathroom scale, if a 140 lb person steps slowly and softly on the scale, the impulse is distributed across a reasonably large interval of time (~1 second), and so the needly just jumps a few pounds past 140 lb. However, if she is a daredevil and decides to leap from the countertop onto the scale (which is probably how most of these household accidents really happen) then the impulse is applied over a very short interval (< 0.1 seconds), and so we can see that the resulting peak dynamic force will be an order of magnitude greater or more. If she squats down on the scale and then jumps up, the force exerted by the legs against the inertia of her body will be added to the acceleration due to gravity she experiences when static; the high she tries to jump (i.e. the more impulse she applies from her legs) the greater the resulting force.
As for the leg pegs on the bike, don’t worry about it. Not only do designers account for the maximum weight of the rider times some reasonable dynamic factor (the 1.5 to 2.0 cited by Santo Rugger are typical dynamic factors for human-powered machinery or slow moving equipment), but then they add on more margin for resistance to fatigue. I would estimate that those foot pegs can accept a load in excess of five times the maximum expected static load. The concern about putting too much load on one side of the bike is somewhat more founded, but remember that the foot peg where this force is being applied is down fairly low and fairly close to the bike centerline (I’m assuming the bike is a cruiser or standard frame by the description of saddle and tailbags), while the rider sits much higher and has his legs outboard. In terms of rolling the bike over, you would sum the moments (the force times the distance from the pivot, which in this case is the line on the ground that would be drawn in-line with the wheels); the moment arm of the force on the peg is much smaller than the moment arm along which the rider is able to act with his legs, so unless you have a golden leg the resulting moment is pretty small for the rider. This is the same reason a rider can hold up a several hundred pound cruise with ease, but if he dumps it he’s going to need a couple of buddies to help him pick it up.
Stranger