Finding a Formula for Determining G Forces in Space Travel

[freespace2dotcom]

Hello.

I'm currently looking for a formula that would determine the G forces that a spaceship (and the people in it) would suffer whilst in it's travels.

The current formula being used takes acceleration of the ship into account, but not a planet's. Therefore, even though I'm supposed to be in a freefall, the formula says I'm experiencing XX G's, even though I shouldn't be.

I'm just not smart enough to figure this out. can anyone help me please?

 

[remcook]

the g you're feeling is basically something pushing you where you don't want to go.
If you stand on the ground, you are accelerated by gravity, but the ground is 'pushing' you up just enough to make you stand still (this sounds strange and maybe it is a strange way of explaining it)

So if you are in your rocket and it is not firing, you and the rocket will both fall towards the Earth (accelerated by gravity). There's nothing keeping you from accelerating freely. But when the rocket fires, it will be accelerated extra. You are basically still only attracted by gravity.

the g will be the difference vector between the Earth gravity and any other acceleration

...I think

need some more time to think about it but have to go. Maybe smeone else can help.

 

[freespace2dotcom]

No.. It's not the difference.

While in an high-ecc. orbit, you accelerate and decelerate gradually. this can go from almost nothing to an acceleration of over 100m/s^2 depending on the characteristics of your orbit. yet during all of this, you feel absoutely nothing but freefall. gravity is a constant that really doesn't change much. the the difference between the two would only work if the acceleration you are currently experiencing is the same as the gravity constant.

 

[da_willem]

the 'g-forces' you feel in a spaceship is just the acceleration of the spaceship in units of g=9,81 m\s^2 minus the gravitational acceleration (gravitational force divided by the mass (m) of the spaceship). In a homogeneous gravitational field this gravitational force is F=mg' with g' the gravitational acceleration. On the Earth's surface g'=g=9,81m/s^2. In a gravitational field due to a point mass (M) or a spherical symmetric mass distribution g'=GM/r^2 with G the gravitational constant 6,6726E-11 Nm^2kg^-2 and r the distance from the center of mass of the mass distribution. (For M=mass of the Earth and r the radius of the Earth g' is just g again).

 

[da_willem]

For circular motion in an orbit around the Earth the acceleration is v^2/r in the direction of the earth. This equals exactly the gravitational acceleration at that positition: GM/r^2 by Keplers law. Both are in the same direction so by the rule I gave you this gives zero g-forces. (you're weightless in orbit)

 

[remcook]

No.. It's not the difference.

While in an high-ecc. orbit, you accelerate and decelerate gradually. this can go from almost nothing to an acceleration of over 100m/s^2 depending on the characteristics of your orbit. yet during all of this, you feel absoutely nothing but freefall. gravity is a constant that really doesn't change much. the the difference between the two would only work if the acceleration you are currently experiencing is the same as the gravity constant.

But the Earth's gravity is still the only thing accelerating you (assuming a Kepler orbit). So the difference between your acceleration and the acceleration by gravity is zero.

 

[remcook]

a point of general discussion (something which I can't get my head around at this moment):

what makes gravity so special that you don't feel it?

ah, just thought of it: Gravity Mass = Inertia Mass
(http://members.tripod.com/~jimmar/index-L5.html)

 

[freespace2dotcom]

Right... I think I see what you're saying.

But please explain the equation in more detail.

Nm^2kg^-2

What is "N" here?

 

[remcook]

it's a unit. N is Newton, the unit of force.

 

[CharlesP]

Be careful to distinguish the g forces you "feel" from the acceleration which happens to you.
The only g force you will feel is the rocket motor pushing you. However much that is and whatever direction it is pointed. When the rocket motor is off you feel nothing (in space).

The acceleration component which is due to gravity (the sum of planets and stars) is not felt. That is the F=GM/r^2. And you can sum over all the universe that you care to think of, to find out what it is.

 

[freespace2dotcom]

"Be careful to distinguish the g forces you "feel" from the acceleration which happens to you. The only g force you will feel is the rocket motor pushing you."

Well, the problem being faced is the need to be able to calulate any g forces that the thing may endure. from orbiting, to atmospheric reentry, etc, etc. If it was just g forces during engine thrust alone, I could ignore g's until the engines were fired up.

 

[da_willem]

The acceleration component which is due to gravity (the sum of planets and stars) is not felt. That is the F=GM/r^2. And you can sum over all the universe that you care to think of, to find out what it is.

So it does not 'feel; different being in space or on the surface of the earth?! It does, and that is due to the acceleration component due to gravity. In the total g-forces you feel you have to take these into account by subtracting them from you acceleration. That's why you don't feel gravity in free fall: your acceleration is equal to the gravitational aceleration (g), so this adds (subtracts) up to nothing.

 

[CharlesP]

So it does not 'feel; different being in space or on the surface of the earth?! It does, and that is due to the acceleration component due to gravity. In the total g-forces you feel you have to take these into account by subtracting them from you acceleration. That's why you don't feel gravity in free fall: your acceleration is equal to the gravitational aceleration (g), so this adds (subtracts) up to nothing.

Lets try one more time. You do not feel the pull of gravity so long as you are free to fall where it takes you. What you feel on Earth is the ground pushing up. At least this is the interpretation of the equivalence principle.