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23 Nov 2019

87.
The force exerted by Earth on a particle of mass madistance r (r > RE) from the center of the Earthhasthe magnitudeofmgRE2/r2, where g=GME/RE2.
 a)Calculatethe work you must do to move the particle fromdistancer1 to r2.
b) Show that when r1 = RE and r2=RE + h, the result can be written as W=mgRE2[(1/RE) - 1/(RE+h)].
c) Show that when h << RE, the work givenisapproximated by W = mgh.

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Related textbook solutions

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

Related questions

Assume the Earth is spherical (radius R) and has uniform
density (mass M). A tunnel is now dug along any secant. A
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opening and dropped from rest. Assume:
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· The tunnel is so thin it has no measurable effect on the
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a) Use Gauss’ law to find an expression for F(r), themagnitude
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center of the Earth, where r < R. Now, find FT(r), the(tangential) component of this force along
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b) Show that m will move through the Earth in simple harmonicmotion. Find T, the period of that
motion. Use: M Earth = 5.98x1024 kg, R Earth = 6.37x106 m, and G =6.67x10-11 N-m2/kg2 to get a
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1. Mail Chute Through the Earth (40 points total)
Assume the Earth is sphere of radius R, has uniform density, massM,
and tunnel dug through the Earth from pole to pole. A box of massm
(very well wrapped!) is held just above the North Pole openingand
dropped from rest (see diagram). Assume:
· You can ignore friction and all effects due to the various
motions of the Earth
· The tunnel is so thin it has no measurable effect on the
gravitational field of the Earth.
a) Use Gauss’ law to find an expression for F(r), the magnitude ofthe gravitational force felt by mass
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b) Show that m will move through the Earth in simple harmonicmotion. Find T, the period of that
motion. Use: MEarth = 5.98x1024 kg, REarth = 6.37x106 m, and G =6.67x10-11 N-m2/kg2 to get a
numerical answer. (8 points)
c) Use a computer program to draw a graph of a(r), (magnitude only,so it is positive at all points),
showing points from r = 0 to r = 2R outside the Earth). (10points)
NOTE: You do not have to show a derivation for F(r) for r >R.
d) Find a numerical value for vmax, the maximum speed of m, usingthe values for the constants given in
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HINT: One method is to use conservation of energy.

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

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