Question for the resident boffins.....

**Jamezo** · 27th February 2006, 09:32

Almost, except that force is a momentary value.

Applying 1N of force to a 1kg object for 1 second results in a kinetic energy transfer of 0.5J (Joules) (Energy = Force * distance, Distance = average velocity * time, in this case 0.5 ms^-1 * 1s = 0.5m. 1N * 0.5m = 0.5J).

So it takes 0.5J to accelerate a 1kg object at 1ms^2 for 1 second.

It takes 1N to accelerate a 1Kg object at 1ms^2, full stop (ignoring relativistic effects lol).

**TwoSeven** · 27th February 2006, 18:43

So the answer to my question was ???

**MSTRS:Me** · 27th February 2006, 18:47

If this was the question....

Originally Posted by TwoSeven

It seems to me that an object travelling at speed away from the earth would be getting lighter - or have I gotten this wrong ?

...then the answer is 'Yes'

**sAsLEX** · 27th February 2006, 19:11

Originally Posted by Jamezo

It takes 1N to accelerate a 1Kg object at 1ms^2, full stop (ignoring relativistic effects lol).

true dont know how I let that second slip in there, had only just got up so a little sleepy

**TwoSeven** · 27th February 2006, 19:18

I suspect that there is a third force in the equation as well. The horizontal force I think needs to be balanced by the lean angle of the machine.

**Hitcher** · 27th February 2006, 19:26

Does the light in the fridge go out when you shut the door?

**Jamezo** · 27th February 2006, 21:19

If you were referring to 'does a motorcycle (or indeed any wheeled contrivance) weigh less as it travels rapidly around the curvature of the earth', then, yes, the answer is yes.

Ok, just for kicks, I shall figure out exactly how substantial this effect might be (the answer, which is rather obvious, is 'not very')

Let us compare the accelerations of the normal gravity force, and the 'driving around the earth very fast' acceleration:

A(G) = 9.81 ms-2, fairly standard

A(D) = v^2 / r, bog standard rotational motion formula

We can take the radius as 6,378m, the equitorial radius of the Earth.

Let us take the speed, for example, as 50ms^-1, or 180km/h.

So the acceleration towards the centre of the earth caused by a motorcycle travelling over the curvature of the earth at sea level and 180km/h can be calculated as:

A(D) = v^2 / r

= 50^2 / 6,378,000
= 0.00039197 ms^-2

So the change is about 0.004% of the total acceleration force of the bike.

Interestingly enough, the orbital velocity at sea level (the point at which these accelerations balance perfectly, and the ground falls away at the same rate gravity attracts it :O), works out to be:

A(G) = A(D)

9.81 = v^2 / r

v = sqr-rt(9.81 * r)
= sqr-rt(9.81 * 6,378,000)

= 7910ms^-1, or roughly 28,476km/h.

Better start fitting that Hayabusa turbo now eh?

**sAsLEX** · 27th February 2006, 21:44

Originally Posted by Jamezo

Interestingly enough, the orbital velocity at sea level (the point at which these accelerations balance perfectly, and the ground falls away at the same rate gravity attracts it :O), works out to be:

or roughly 28,476km/h.

but escape velocity is
v = sqrt(2gR)
The value evaluates to be approximately:

11100 m/s
40200 km/h
25000 mi/h

**Virago** · 27th February 2006, 22:06

Originally Posted by Jamezo

.......Interestingly enough, the orbital velocity at sea level (the point at which these accelerations balance perfectly, and the ground falls away at the same rate gravity attracts it :O), works out to be:

A(G) = A(D)

9.81 = v^2 / r

v = sqr-rt(9.81 * r)
= sqr-rt(9.81 * 6,378,000)

= 7910ms^-1, or roughly 28,476km/h......

And to think I took a rough guess and said 30,000km/h - how wrong can a person get?

**scumdog** · 28th February 2006, 00:24

Now as you know for a fixed number of revolutions a smaller diameter wheel travels slower than a big one - covers less ground for a given number of turns.
So does that mean you go slow-fast-slow as you go through S bends - if you keep the same revs?

After all the centre of the tyre is in effect larger wheel than the shoulder of same tyre....

**MSTRS:Me** · 28th February 2006, 09:23

Thinking about this, I come up with.....
If you are a motorcyclist who does NOT accelerate through and out of a corner then one of two things will be the case.....you will run wide or you ride a Harley (or both)

**Ixion** · 28th February 2006, 09:25

Motards ? .

**scumdog** · 28th February 2006, 09:44

Originally Posted by MSTRS

Thinking about this, I come up with.....
If you are a motorcyclist who does NOT accelerate through and out of a corner then one of two things will be the case.....you will run wide or you ride a Harley (or both)

Ah, but how do you KNOW that your accelerating and not just keeping the same speed and the revs go up to compensate for the 'reduction in tyre diameter??

And by your theory if I stop braking while going around corners I would run off the INSIDE of the bend???

**Jamezo** · 28th February 2006, 10:42

Originally Posted by sAsLEX

but escape velocity is
v = sqrt(2gR)
The value evaluates to be approximately:

11100 m/s
40200 km/h
25000 mi/h

nah, escape velocity is a hypothetical initial vertical velocity which would enable you to escape completely from the earths gravitational acceleration, to a hypothetical point an infinite distance from the earth, but in reality, just so far away that the inverse square relation of distance to gravitational acceleration renders it irrelevant.

what I calculated was in essence the orbital velocity at sea level, quite a different kettle of fish.