Gravitational Energy in Planetary Motion, Assignments of Physics

The concept of gravitational energy in the context of planetary motion. It discusses how the total energy of a planet in circular or elliptical orbit around a center of force is given by eq. 14-25, which includes both kinetic and potential energy. The document also explains how the total energy determines the semimajor axis and period of the orbit, and how changing the speed of an orbiting satellite requires a change in radius. A sample problem is provided to illustrate the concepts.

Typology: Assignments

Pre 2010

Uploaded on 08/31/2009

koofers-user-kgz
koofers-user-kgz 🇺🇸

9 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
~)
rv-;,L:; i
Jv...
VZ-
v:::
V
2,L
:::
It,,r
~
(}1)
.J:.~~
~.~
~
U1~
tr--
d
(1
r )
\(")
":/
\..
............
v~t-
wf--
IT'f
.
~
'c-u
U
~
f.t--~
(Lr)
'0/~~
~~
/L=tfV'il2-
U::~~+t~)
v""--
')
CL
r-
)
~
~
kr
IJ
+--
~J
(I
t-rCA-..(y")
7-
~
(?-!
'I
~
{f~/-
~(9
4-;7
(y/
tr1~)
2
~
i
~(9
-I-
!
-I-
~
1
-:.
~
<A...
r9-
G---t2
::
r
(y..
~
«0.
l
c)
pf3
pf4

Partial preview of the text

Download Gravitational Energy in Planetary Motion and more Assignments Physics in PDF only on Docsity!

~) rv-;,L:; i Jv... VZ-

v::: V2,L ::: It,,r ~

(}1) .J:.~~ ~.~ ~

U1~ tr-- d (1 r )

(") ":/ .. ............ v~t-

wf-- IT'f. ~ 'c-u U ~ f.t--~ (Lr)

'0/~~ ~~ /L=tfV'il2- U::~~+t~)

v""-- ') C L r- ) ~ ~ kr IJ 7 +-- ~J (I t-rCA-..(y")

~ (?-! 'I ~ {f~/- ~(9 4-;7(y/ tr1~)

2 ~ i ~(9 -I- !-I- ~

1 -:. ~ <A... r9- G---t2 :: r (y.. ~ «0. l c)

/lAvl

- r-

M :::~2-

r

()~)r

vf- "': Ie ~u U: fIAj J--, ~

l

tet·~ ~ ~ lL:: f rv1/ U -- Y1A j (2,- )

-t/VL^ +- ~ (14') .~. rj h

1 CXr + 1- d^ r '-. ft!t.., A:: 2,)r

~/l 1 I

/ )^ J.",^ tty )^ ~ .¢ e..j^ 1.^ /Ij

uL..-~ tv j r

~ = "'1 +- ~l =:"'1 + ~ LDez/ ::::~)

u z: ~ (w ) ~ ::: { wvl

~ (Lr/ -+ 1 wV-l.--:;. ~ (~r)

if,,"1--": Y>(5r) v

L

,:::- 0Jr

PN~? ::- v:;; ~ ~. ~;-- ~ ~?- i?'Hr^ - ~ i -:: S

14-8 THE GRAVITATIONAL FIELD (OPTIONAL) 315

FIGURE 14-19. Sample Problem 14-10. The orbits of space

craft A and B are shown. Note that B catches A by moving to a

noncircular orbit at lower height above the Earth. The relative size of the Earth and the orbital heights is not to scale.

After the burn, the speed decreases by the given amount of 0.95%

to v' (l - O.OO95)v = 7.68 X 103 mIs, and the new kinetic en
ergy of B is
K' ~(3250 kg)(7.68 X 103 m/s)2 = 9.58 X 1010 1.

The potential energy of B at point P immediately after the short

burn is unchanged, equal to the initial value E K or 2E, accord
ing to Eq. 14-25. The total energy E' of B after the burn must then

be E' K' + U' = 9.58 X 1010 J + 2(- 9.76 X 1010 J)

  • 9.94 X 1010 J,
and the new semimajor axis is, from Eq. 14-25,
_ GmM E
a' 2Et
(6.67 X 10- 11 N· m^2 /kg 2 )(3250 kg)(5.98 X 1024 kg)

= ......... 2(-9.94 X lOiO J)

6.52 X 106 m 6520 km, a reduction of 1.8% from the value in the original orbit. The corre sponding period is

41T2at3 )
T' (
GM E

417'2(6.52 X 106 mi ) == ((6.67 X 10- 11 N' m^2 /kg^2 )(5.98 X 10 H^ kg)

'" 5240 s.

The difference in the periods is 140 s. That is, if A originally
passes through point P at t = 0 and B passes through (and fires its

rockets) at t = 105 s, then A returns to P at t 5380 s (deter

mined by the period T), and B returns to P at 5240 s after its ini

tial passage, or at t = 105 s + 5240 s 5345 s. Thus B is now

35 s ahead of A at point P. Now B can fire a second rocket burst

identical in strength and duration to the first but in the reverse di rection. This returns B to the original circular orbit, now 35 s

ahead of A. Figure 14-19 shows the relationship between A and B
during the first orbit after the bum. Note that after the burn, B
moves in an elliptical orbit and so can pass A without colliding be
cause A remains in the original circular orbit.
See Exercise 38 to help understand how B can reduce its speed
at P and still get ahead of A.

1 4-8 THE GRAVITATIONAL FIELD (Optional)

A basic fact of gravitation is that two particles exert forces on one another. We can think of this as a direct interaction between the two particles, if we wish. This point of view is called action-at-a-distance, the particles interacting even though they are not in contact. Another point of view is the field concept, which regards a particle as modifying the space around it in some way and setting up a gravitational field. This field, the strength of which depends on the mass of the particle, then acts on any other particle, exerting the force of gravitational attraction on it. The field therefore plays an intermediate role in our thinking about the force that one particle exerts on another. According to this view we have two separate parts to our problem. First, we must determine the gravitational ficld established by a given distribution of particles. Sec ond, we must calculate the gravitational force that this field exerts on another particle placed in it. We use this same approach later in the text when we study electromagnetism, in which case particles with elec tric charge set up an electric field, and the force on another charged particle is determined by the strength of the electric field at the location of the particle. Let us consider the Earth as an isolated particle and ig nore all rotational and other nongravitational effects (so that g and go are equivalent). We use a small test body of mass

mo as a probe of the gravitational field. If this body is

placed in the vicinity of the Earth, it will experience a force having a definite direction and magnitude at each point in space. The direction is radially in toward the center of the Earth, and the magnitude is mog. We can associate with each point near the Earth a vector g, which is the accelera tion that a body would experienee if it were released at this point. We define the gravitational field strength at a point as the gravitational force per unit mass at that point or, in terms of our test mass,

F g= (14-26)

By moving the test mass to various positions, we can make a map showing the gravitational field at any point in space. We can then find the force on a particle at any point in that field by multiplying the mass m of the particle by the value of the gravitational field g at that point: F mg. Figure 14-20 shows examples of gravitational fields. The gravitational field is an example of a vector field, each point in this field having a vector associated with it. There are also scalar fields, such as the temperature field in a heat-conducting solid. The gravitational field arising from

-