Recent content by Raphael30

From what I see everywhere, the way people solve this is: 1) use conservation of energy to find an equation for the remaining relative velocity as the distance from the Earth tends towards infinity for any burnout velocity greater then escape velocity 2) invert this to have the burnout velocity...

That's what Wikipedia says at least: https://en.wikipedia.org/wiki/Hohmann_transfer_orbit. Too bad they don't explain how to get such low Δv.

My attempt so far: The difference in energy between the Earth's revolution orbit and the transfer orbit is μS(RMarsRev-(REarthRev)/(2REarthRev(RMarsRev+REarthRev)). If we express the satellite's energy as the sum of the Earth's and the satellite's, the difference between the satellite's energy...

Homework Statement We want to send a satellite from a low Earth orbit of 320 km to mars. Calculate the change in velocity required to join the transfer ellipse. Homework Equations Earth velocity: (μS/REarthRev)1/2 Transfer velocity at perihelion: (2μSRMarsRev/(REarthRev(REarthRev+RMarsRev))1/2...

I don't see the the difference with the way we previously solved it. The only possible mistake I could see at this point, which would make sense, is the hypothesis that the free end is free falling, which gave us the expression for v(y). If v(y) didn't drop at 0 suddenly when y=L, neither would...

What do B and R stand for here?

Yes, I noticed that didn't work. Also, dp/dy doesn't equal vdm/dy, dp/dy=vdm/dy+m(y)dv/dy. This doesn't affect the end result, dp/dy now simply incorporates Fg. The end result should therefore be R=dp/dt=(mg/2)(1+3y/L) until y=L, where R falls from 2mg to Fg=mg.

Ok so my idea is F=dp/dt=(dp/dy)(dy/dt) dy/dt=v=(2yg)1/2 dp=dm x v=(my/2L)(2gy)1/2dy therefore F=(my/2L)(2yg)1/2(2gy)1/2=gmy2/L where R=Fg+F=(mg/2L)(L+y+2y2) until y=L, where R falls to Fg=mg I think this makes sense, doesn't it?

Homework Statement A rope of length L and mass m is attached to the ceiling and folded at mid length. The free end of the rope is held against the ceiling, then dropped at time t=0. a) Calculate the velocity of the free end as a function of y, its distance from the ceiling. b) Calculate R...

Thanks! I have one final question: in this case, how does the collision affect the tension in both ropes immediately after? The added mass of the ball should increase the tension in the right rope, but is there another factor when the plate is still straight? I was thinking maybe the centripetal...

Made this beautiful paint drawing. The pink line is where I suppose the resulting instantaneous axis of rotation should be, which is why I said ω has x and y components. Since gravity and tension are both on the same plane as the axis of rotation at this instant (and I only have to calculate...

For a question involving a ball hitting orthogonally the bottom corner of a board held by wires, I need to calculate the angular velocity of the board and ball (collision is inelastic) right after the collision, before there's any external torque. I can easily calculate the angular momentum L...