Solve Number 2 in Free Space: a=(Vrk/(1-kt))

[mopar969]

Here is the problem: In free space, what would the answer to number 2 be?
Number 2 is a=((Vrk)/(1-kt))-g.

I think that since there is no gravity in free space the answer is a=((Vrk)/(1-kt)). Is this correct and is my reason for it correct?

 

[Delphi51]

What is the question?
There IS gravity in space. It is the gravitational force that holds the moon in orbit around the Earth.

 

[Stonebridge]

Maybe.
Could you explain the formula and what the various terms represent?

 

[mopar969]

This is the formula for the acceleration of a rocket. I thought that since the problem is in free space there is no gravity? Please help me figure out the equation? Also another note is that the rate of ejection of mass by the rocket is (dm/dt)=-km(initial). I used this and substituted it into the original acceleration equation I had a=(-vr/m)(dm/dt)-g.

 

[Delphi51]

It is very difficult for us to give useful help unless you show us the entire question, word for word.

 

[mopar969]

The entiree problem word for word is :
The acceleration of a rocket fired vertically upward is a = (-Vr/m)(dm/dt)-g. Suppose the rate of ejection mass by a rocket is a constant (dm/dt)=-km(initial) therefore, m=m(initial)(1-kt). Therefore under the conditions of the previous equation a = ((Vrk)/(1-kt))-g. Now, in free space, what would the answer to the last equation be?

 

[Delphi51]

Okay, it makes sense now. It made sense to subtract g when the rocket was taking off from the surface of the Earth, but not once it is in space. If there is a gravitational acceleration (as in near Earth or the moon) it wouldn't be precisely opposite the direction of thrust anyway. Dropping the g looks good.