Question

2) A satellite of mass m is in circular orbit of radius ro in a potential of the form V(r) --kr-3/2 (a) Use the Viral Theorem to find the orbital speed of the satellite. (b) What is the value of the one-dimensional effective potential at ro, Vefr(ro), for this situation? (c) Sketch Velf as a function of r for this potential, labeling your axes carefully, The satellite is now given an impulse which increases its kinetic energy by 20%. (d) On a copy of the effective potential from (c), sketch the new Veff if the impulse is radial. Qualitatively indicate on your plot the new total energy of the satellite and the apsidal distances (turning points) of the new orbit. Your plot should clearly indicate how the effective potential has changed between parts (c) and (d). (e) Repeat part (d) if the impulse is angular. Again your plot should indicate the difference between (C) and (c).

Answer 1

a) The Viral Theorem indicates the following:

(1)

«T>= Σ< F, ή >

Where

< T>: is the average kinetic energy on k particles

vec{F_{k}: is the force of attraction between the particles

vec{r_{k}}:is the radial distance between the particles

In our case, we have one particle, then

(2)

T=-*<Ēr>

The radial force exerted on the satellite is expressed as a function of the potential

(3)

(1) 4-=

Where

(4)

+1+z/5-145 = "ZAP = 14

In magnitude

(5)

ا = سنت انا - s/

but the satellite position vector and the radial force are opposite, therefore when making the dot product between these vectors, the angle is 180 degrees

T=-*<Ēr>

(6)

T = Firicos180

$T = \frac{3}{2}kr^{-5/2}r=\frac{3}{2}kr^{-3/2}$ (7)

Now

(8)

T= -01 m

V =

Therefore

3 v=1-kr-2 V 2m

b)The effective potential is defined as the potential due to the radial force, plus the potential due to the centrifugal force, whose force moves the satellite away from the mass that attracts it

(9)

Vefs = V(r) +m!

Where:

l: is the angular momentum

Vefp = -kr-3/2+ 2r2

Adding the fractions of the previous equation we have:

(10)

Vefs = -2krém12,3/2 2r7/2

c)

For the realization of the previous graph we assume from equation 10, the following:

$l=1$

m = 1

$k=1$

(11)

-2r2 + 3/2 Veff = 277/2

when graphing expression 11, we have:

0 2 4 6 8 10 12 Eeff radius