Question

A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r2 because A = πr2. Determine how the following properties of the satellite depend on r. In each case, give the power of r, i.e., if the dependence goes as rn, give n. If there is no dependence, then n = 0. (a) period, (b) kinetic energy, (c) angular momentum, and (d) speed.

Answer 1

Answer:

a. $T=r^{3/2}$

b. $K=\frac{1}{r}$

c. $v=\frac{1}{\sqrt{r}}$

d. $v=\sqrt{r}$

Explanation:

To make analysis about the satellite circular earth the depends or r and A

$T^2=\frac{4\pi}{GM}*r^3$

$K=\frac{GM*m}{2*r}$

a.

$T^2=r^3$

$T=r^{3/2}$

b.

$K=\frac{GM*m}{2}*\frac{1}{r}$

$K=\frac{1}{r}$

c.

$K=\frac{1}{2}*m*v^2$

$v^2=\frac{2*K}{m}=\frac{2*Gm*m}{2*m*r}$

$v=\frac{1}{\sqrt{r}}$

d.

$v=\sqrt{2*GMr}$

$v=\sqrt{r}$

Answer 2

The way that the period, kinetic energy, angular momentum, and speed of the satellite depend on r are; 1) T ∝ r^(³/₂) 2) K.E ∝ 1/r3) L ∝ √r 4) v ∝ 1/√r

What are the physical properties of the satellite?

A) For circular motion, centripetal force is balanced by gravitational force. Thus;

F_centripetal = F_grav

mrω² = GM_earth*m/r²

⇒ ω² = GM_earth/r³

ω = 2π/T. Thus;

(2π)²/T² = GM_earth/r³

T² = ((2π)²/GM_earth)r³

(2π)²/GM_earth is a constant and as such we are left with;

T ∝ r^(³/₂)

B) Kinetic energy is expressed as;

K.E = ¹/₂mv²

Thus;

¹/₂mv² = GM_earth*m/r²

v² = 2GM_earth/r²

Reduces to;

v ∝ 1/√r

Since K.E ∝ v², then;

K.E ∝ 1/r

c) For the angular momentum, we have;

L = Iω

L ∝ √r

D) From B above we saw that for the speed, we have the expression;

v ∝ 1/√r

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