How is Rocket Propulsion Affected by Gravity?

AI Thread Summary
Rocket propulsion is significantly affected by gravity when launched from the Earth's surface, as it introduces an additional force that must be considered alongside the thrust generated by the rocket's engines. The modification of equation (8.37) requires incorporating the gravitational force, leading to the expression m(dv/dt) + mg = -v(dm/dt). This adjustment reflects that the net external force includes both the thrust and the weight of the rocket. To derive the acceleration for part (b), one can rearrange the modified equation to solve for dv/dt. Understanding these dynamics is crucial for accurately modeling rocket motion in a gravitational field.
Lola Luck
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Homework Statement



Earlier we considered a rocket fired in outer space where there is no air resistance and where gravity is negligible. Suppose instead that the rocket is accelerating vertically upward from rest on the Earth's surface. Continue to ignore air resistance and consider only that part of the motion where the altitude of the rocket is small so that g may be assumed to be constant.

a) How is eq. (8.37) modified in the presence of the gravity force?

b) Derive an expression for the acceleration a of the rocket, analogous to Eq. (8.39)

Homework Equations



eq (8.37): m(dv/dt) = - vex(dm/dt)
m= mass of rocket vex= velocity of rocket exhaust v= velocity of rocket

eq (8.39): a= (dv/dt) = (-vex/m)(dm/dt)

The Attempt at a Solution



For part (a) I believe I need to factor in the impulse of gravity, Jg= mg(dt), but I don't know how. Then I would use my answer to part (a) to find an equation for acceleration.

Thank you
 
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Lola Luck said:
For part (a) I believe I need to factor in the impulse of gravity, Jg= mg(dt), but I don't know how.
You need to include the effects of gravity, but not as you've stated. Try looking at units for 8.37 and rethink what you "believe."
 
i believe that the trick lies in realising the fact that the net external force acting on the rocket is now, no longer the reaction due to its own engines, but also gravity acts on it. so we need to modify the LHS first
 
sorry for the late response!

so the units of the equation are kg*(m/s^2), which is the same as a Newton. So do I just add the force of gravity?

m (dv/dt)+mg=-v(dm/dt)
 
That'd be my first approach.
 
okay, and then I just solve for dv/dt to get acceleration for part (b). thank you!
 
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