Rocket Propulsion

mopar969

15th November 2010 - 01:44 AM

Suppose the rate of ejection mass by a rocket is a constant, dm/dt = -kmo. I need to show that m=mo(1-kt). I know -mg=m(dv/dt)-Vr(dm/dt). I thought that I could just substitute for (dm/dt) and get the solution but I am having no luck. Please help me show this property.

Umer.banday

12th March 2011 - 09:26 AM

Here dm is the small fraction of the released mass...if we have to find the total mass released in time t, we have to add all these small "dm"s together i.e we have to integrate it..nw dm/dt=kM0 OR dm=kM0dt now integrate both sides we get M=kM0t..now.leftover mass is M0-M i.e M0-kM0t..take out M0 then we have M0(1-kt)...

PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.