The exhaust velocity for stage I of a Saturn V rocket is 2.6 km/s. The escape ve

Question

The exhaust velocity for stage I of a Saturn V rocket is 2.6 km/s. The escape velocity of Earth is 11.2 km/s. What's the minimum fraction of rocket + fuel + propellant that must be exhausted to reach this velocity with a Saturn V rocket's first stage One is taught that Newton's second law of motion is F = ma, where F is force, m is mass, and a is acceleration. This isn't actually Newton's second law; it's a common simplification. Newton actually wrote F = p, where p is the momentum, and the overset dot means the derivative with respect to time (this was Newton's notation for derivatives). For a point particle, p = mv, and we would write Newton's second law as , F(t) = d/dt (mv). Use the product rule to rewrite F(t), assuming m and v are functions of time. Remember that we have another name for the rate of change of velocity. Explain what common simplification is made to get F = ma from the more complicated expression you got. When a rocket heats and expels propellant, its mass is reduced and both terms you got in part 1 are relevant for rocket science. With sufficient ingenuity, one can show that integral^t_f_t_0 a(t) dt = -v_e integral^m_f_m_0 1/m dm, where t_0 and t_f are the start and end time of the burn, m_0 and m_f are the start and end masses, and v_e is the exhaust velocity of the propellant (in the frame of reference of the rocket, in a vacuum). Do the integrals. Use delta v = v_f - v_0 for the change in velocity and arrange the masses to appear in the ratio m_0/m_f. If the increase of velocity of a rocket must be twice the exhaust velocity of the propellant, what must the ratio of the initial (rocket + ejected propellant + fuel burned) mass to the final (just rocket) mass be What fraction of the mass of the rocket is burned in this process

Explanation / Answer

Part 1)

Make de derivade

F = d (mv)/dt

F = m dv/dt + v dm/dt

F = m a + v dm/dt

Part 2)

In most situations the physical mass of the bodies is constant for which the second term is zero.

    Even case nuclear reactions of the small amount of mass is lost transform in energy, but in ordinary chemical reactions constant, the only mass changes between the union of the components, for which the second term is zero

In a number of cases this mass changes in bodies, in some cases as rockets and other decreases as the mass increases conveyors, in this case the term second mass is not zero.

Part 3)

a dt = (dv/dt)dt = dv = Vf– Vo

- Ve m-1dm = - Ve(ln mf- ln mo) = - Veln ( mf/mo)

Vf- Vo = - Ve ln ( mf/mo )

Vf- Vo = Ve ln ( mo/mf ) (1)

Part 4)

The problem is expressed

Vf – Vo = 2Ve

Calculate with 1

2Ve = Ve ln ( mo/mf)

2 = ln ( mo/mf )

e2 = mo/mf

( mo/mf) = 7.389

Part 5)

Ve = 2.6 Km/s

Vf = 11.2 Km/s

Vo = 0

We used expression 1

Vf- Vo = Ve ln ( mo/mf )

11.2 – 0 = 2.6 ln ( mo/mf )

11.2/2.6 = ln ( mo/mf )

4.308 = ln ( mo/mf )

e4..308 = ( mo/mf )

( mo/mf) = 74.27

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