Ch 5.3 - 14E

QUESTION:

(a) Use the substitution \(v=d y / d t\) to solve (13) for v in terms of y. Assuming that the velocity of the rocket at burnout is \(v=v_{0}\) and \(y \approx R\) at that instant, show that the approximate value of the constant c of integration is \(c=-g R+\frac{1}{2} v_{0}^{2}\).

(b) Use the solution for v in part (a) to show that the escape velocity of the rocket is given by \(v_{0}=\sqrt{2 g R}\). [Hint: Take \(y \rightarrow \infty\) and assume \(v>0\) for all time t.]

(c) The result in part (b) holds for any body in the Solar System. Use the values \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) and \(R=4000 \mathrm{mi}\) to show that the escape velocity from the Earth is (approximately) \(v_{0}=25,000 \mathrm{mi} / \mathrm{h}\).

(d) Find the escape velocity from the Moon if the acceleration of gravity is 0.165g and \(R=1080 \mathrm{mi}\).

Text Transcription:

v=d y/d t

v=v_0

y\approx R

c=-gR+frac12v_0^2

v_0=sqrt2gR

y\rightarrow\infty

v>0

g=32ftmathrms^2

R=4000mi

v_0=25,000mih

R=1080mi