### Post by PhantomWolf on Feb 22, 2012 17:27:52 GMT -4

I had a nice long post explaining all about why to and not to include gravity, but Explorer crashed, and since Prof is likely to ignore it anyway......

So the abridged version...

F = ma

where a

and g

Now we have two basic situations, one is where our plane is vertical to the local gravity field and so g

So given that...

F = m (a + g

First let's take a hovering spacecraft of mass 2,500kg. Since it is hovering, a = 0m/s

Since we are dealing with the Vertical Plane we know that;

On Earth g

On Moon g

so on Earth...

F = m (a + g

F = 2500kg (0ms

F = 2500kg x 9.81m/s

F = 24,525N

On the Moon

F = m (a + g

F = 2500kg (0ms

F = 2500kg x 1.63m/s

F = 4,075N

Hopefully this expains why you need less thrust on the moon, it's not the mass that has changed, but the acceleration you need to overcome.

Now let's look at a ball moving horizontally in both places. Remember that in the horizontal plane, g

Our ball will have a mass of 0.5kg and we'll throw it with an acceleration of 48m/s

so on Earth...

F = m (a + g

F = 0.5kg (48ms

F = 0.5kg x 48m/s

F = 24N

On the Moon

F = m (a + g

F = 0.5kg (48ms

F = 0.5kg x 48m/s

F = 24N

Note that the answer is identical in both equations. Force that is horizontal to the local gravity vector doesn't change regardless of where it is in the universe.

So how does that effect the energy?

Well E

so regardless of where the horizontal force occurs, the E

In the case of the balls in the above example, 576J

This should also explain why your don't change the density (which is mass/volume) as mass doesn't change, and why you don't consider gravity when dealing with horizontal motion, such as dust being blown across a surface.

ETA: Something I should have noted, and did in the longer version (yes it was longer ) is that while the ball leaves our hand with the same velocity in both examples (since v = at and a and t are the same in both examples) it will go futher on the Moon because after leaving our hand the lower gravity means it will take longer to reach the ground allowing it to travel a much greater distance.)

So the abridged version...

F = ma

_{t}where a

_{t}= a + g_{p}and g

_{p}is the acceleration vector component of gravity in the plane of the Force vector we're looking atNow we have two basic situations, one is where our plane is vertical to the local gravity field and so g

_{p}= our local g and where the other is when the plane is horizontal to the local gravity field and so our g_{p}= 0m/s^{2}regardless of where it is in the universe because gravity is a vector that has no horizontal component to it. (Note that a combination of these two planes can be used to represent any force vector possible which is why they are the simplest planes to work with to demonstrate the point.)So given that...

F = m (a + g

_{p})First let's take a hovering spacecraft of mass 2,500kg. Since it is hovering, a = 0m/s

^{2}Since we are dealing with the Vertical Plane we know that;

On Earth g

_{pv}= 9.81m/s^{2}On Moon g

_{pv}= 1.63m/s^{2}so on Earth...

F = m (a + g

_{pv})F = 2500kg (0ms

^{2}+ 9.81m/s^{2})F = 2500kg x 9.81m/s

^{2}F = 24,525N

On the Moon

F = m (a + g

_{pv})F = 2500kg (0ms

^{2}+ 1.63m/s^{2})F = 2500kg x 1.63m/s

^{2}F = 4,075N

Hopefully this expains why you need less thrust on the moon, it's not the mass that has changed, but the acceleration you need to overcome.

Now let's look at a ball moving horizontally in both places. Remember that in the horizontal plane, g

_{ph}= 0m/s^{2}because gravity has no horizontal component.Our ball will have a mass of 0.5kg and we'll throw it with an acceleration of 48m/s

^{2}taking 1s to do it.so on Earth...

F = m (a + g

_{ph})F = 0.5kg (48ms

^{2}+ 0m/s^{2})F = 0.5kg x 48m/s

^{2}F = 24N

On the Moon

F = m (a + g

_{ph})F = 0.5kg (48ms

^{2}+ 0m/s^{2})F = 0.5kg x 48m/s

^{2}F = 24N

Note that the answer is identical in both equations. Force that is horizontal to the local gravity vector doesn't change regardless of where it is in the universe.

So how does that effect the energy?

Well E

_{k}= ½F.a.t^{2}so regardless of where the horizontal force occurs, the E

_{k}it transmits will be the same.In the case of the balls in the above example, 576J

This should also explain why your don't change the density (which is mass/volume) as mass doesn't change, and why you don't consider gravity when dealing with horizontal motion, such as dust being blown across a surface.

ETA: Something I should have noted, and did in the longer version (yes it was longer ) is that while the ball leaves our hand with the same velocity in both examples (since v = at and a and t are the same in both examples) it will go futher on the Moon because after leaving our hand the lower gravity means it will take longer to reach the ground allowing it to travel a much greater distance.)