PCS 125 Lecture 12: Chapter 23- Chapter Summary

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9 Apr 2016

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Related Questions

Figure 1 illustrates a Keplerian orbit, with Cartesiancoordinates (x,y) and polar coordinates (r,Ï).

Parametric equations for a general Keplerian orbit are r(Ï) = a(1 â e cosÏ)

t(Ï) = *(T)/2Ï+ ( Ïâ e sinÏ)

tan(Ï/2) = *(1+e)/(1âe)+1/2 tan(Ï/2)

Here Ï is an independent variable. The solution is periodic in Ïwith period âÏ = 2Ï. Also, a, e, T are constants that determine theparameters of the orbit.

Figure 2 shows the relation between Ï and the spatialcoordinates. In the figure, the coordinates (Î¾,Î·) are defined by Î¾= x + ae and Î· = y a /b,

withb=a*1âe2 ]1/2.

(A) What kind of curve is the orbit?

(D) Determine Î¾(Ï) and Î·(Ï).

(E) Express r as a function Ï .

(F) The angular momentum is L = m r2dÏ/dt. Determine L in termsof {a,e,T} Hence verify that L is a constant of the motion.

(G) The energy isE = [1/2] m ( dr/dt )2+ [1/2] m r2(dÏ/dt)2â GMm/r.Determine E in terms of {a,e,T}.

(H) E must be a constant of the motion. Hence determine T and Efrom the equation that you obtained in (F).

(I) Write T in terms of the spatial orbit parameters and theforce constant GM.

(J) Now express L and E in terms of a, e, GM.

(K) In Figure 1, determine the coordinates of the points P, A,and R

(L) In Figure 1, determine the time t at the point R.

Figure below shows a cycle undergone by 1.00 mol of an idealmonatomic gas. The temperatures
are T1 = 300 K, T2 = 600 K, and T3 = 455 K.
For 1 ? 2, what are
(a) (2 points) heat Q?
(b) (2 points) The change in internal energy ?Eint?
(c) (2 points) The work done W?
For 2 ? 3, what are
(d) (1 points) Q?
(e) (2 points) ?Eint?
(f) (2 points) W?
For 3 ? 1, what are
(g) (2 points) Q?
(h) (2 points) ?Eint?
(i) (2 points) W?
For the full cycle, what are
(j) (2 points) Q?
(k) (2 points) ?Eint?
(l) (2 points) W?
The initial pressure at point 1 is 1.00 atm (= 1.013 Ã 105 Pa).What are the
(m) (2 points) volume?
(n) (2 points) Pressure at point 2?
(o) (2 points) Volume?
(p) (1 points) Pressure at point 3?

aquamarinegnat75

A particle of mass m moves under the influence of a central force.The potential energy function is
V(r) = {
K (a2âr2) /2 if r â¤ a
0 if r > a

as illustrated in the figure. The initial position is (x,y) = (a/2,0) and the initial velocity is (vx,vy) = (0, v0).
[Data: m = 1.0 kg; K =17.1 N/m; a = 1.0 m; r0 = 0.5 a; v0 = 2.0m/s.]

(A) Determine the time ta when the mass crosses the boundary r =a.

(B) Determine angular position Ïa where the particle crosses theboundary r = a.

***HINT GIVEN***

Use the constants of the motion to analyze the orbit.

Write L = mr^2 dÏ/dt = the initial value.
What is the initial value?

Write E = (1/2)m (dr/dt)^2 + L^2/(2mr2) +V(r )
= the initial value. What is the
initial value?

Part A. In the energy equation,
change the variable from r to u,
where r =sqrt(u). The result can be
solved by u(t) = c1+c2 cosh(Î±t).
Determine c1, c2, and Î±.

Part B.
Ï = Integral 0 to t [L/(mr^2)] dt

PCS 125 Lecture 12: Chapter 23- Chapter Summary

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