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PHYS 204A Midterm: PHYS 204A Chico State Exam1 eqsold

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turquoisegnu128
31 Jan 2019
School
Chico State University
Department
Physics
Course
PHYS 204A
Professor
All

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Document Summary

The definition of velocity: r v ! r r d dt d r v dt. The kinematic equations: v = v: + at x = x, + v t + 1. 2 v 2 = v o x ! x o = 1. Ab cos" = axbx + ayby + azbz. Acceleration due to gravity g = 9. 80 m/s2. Earth - mass: 5. 98 x 1024 kg radius: 6. 38 x 106 m. Moon - mass: 7. 36 x 1022 kg radius: 1. 74 x 106 m. Sun - mass: 1. 99 x 1030 kg radius: 6. 96 x 108 m. Earth - moon distance: 3. 82 x 108 m.

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Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

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Related Questions

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

Parametric equations for a general Keplerian orbit are r(ψ) = a(1 − e cosψ)

t(ψ) = *(T)/2π+ ( ψ− e sinψ)

tan(φ/2) = *(1+e)/(1−e)+1/2 tan(ψ/2)

Here ψ is an independent variable. The solution is periodic in ψwith period ∆ψ = 2π. Also, a, e, T are constants that determine theparameters of the orbit.

Figure 2 shows the relation between ψ and the spatialcoordinates. In the figure, the coordinates (ξ,η) are defined by ξ= x + ae and η = y a /b,

withb=a*1−e2 ]1/2.

(A) What kind of curve is the orbit?

(C) Determine x(ψ) and y(ψ).

(D) Determine ξ(ψ) and η(ψ).

(E) Express r as a function φ .

(F) The angular momentum is L = m r2dφ/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(dφ/dt)2− 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.

Consider a satellite of mass ms in a circular earth orbit. Anysatellite orbiting in the upper reaches of our atmosphere willexperience a small drag force, Fd, opposite to itsdirection of motion. This drag force will cause the satellite tospiral inward toward smaller orbital radii. In the process, theorbital velocity of the satellite actually increases! This problemwill investigate how a drag force can cause a body in orbit tospeed up.

The situation is sketched in Fig. 1. The diagram shows separateforce and velocity diagrams for the satellite. The mass of theearth is denoted me, while the mass of the satelliteis ms. The tangential velocity of the satellite isvt and the small radial velocity is acquires because itis spiraling inwards is vr. The size of vr isgreatly exaggerated in the diagram, and we can take vr<< vt and use vt as an adequateapproximation to the magnitude of the total velocity, v. We willcall θ the ‘angle of de- scent’.

Turning to the force diagram, we see that the total force actingon the satellite is the vector sum of F⃗d andF⃗g. Again, the drag force is much weaker than thegravitational force, so Fd << Fg . Wecan use this to our advantage by just considering the effect ofFg along the line of descent. We can see from thediagram that Fg has a component along the line ofdescent given by Fg sin(θ).

(a) First show how we arrive at the following two basicequations governing this system, and carefully describe thephysical meaning of the quantity msad inwords:

Fgsinθ−Fd = msad (1)

Fg = (ms*vt2)/r= G(msme)/r2 (2)

(b) We can find out more about ad and its relation tothe various forces and velocities by powerful

use of the chain rule for differentiation:

Explain Eq. 3 in your own words. Where did vr comefrom and why is the ≃ sign there?

ad = dv/dt = (dv/dr)*(dr/dt) =vr(dv/dr) ≃ vr(dvt/dr) (3)

(c) We can now make use of Eqs. 2 and 3 to show that:

msad = (1/2)Fg sinθ