Question

i) A satellite revolves in a circular orbit around the earth at a certain height (h) above
it. Suppose that h << R, the radius of the earth. Calculate the time period of
revolution of the satellite.
ii) Derive expressions for average energy of a body executing SHM.

Answer 1

Answer on Question #61824, Physics / Other

i) A satellite revolves in a circular orbit around the earth at a certain height (h) above it. Suppose that $h \ll R$ , the radius of the earth. Calculate the time period of revolution of the satellite.

ii) Derive expressions for average energy of a body executing SHM.

Solution:

(i)

This net centripetal force is the result of the gravitational force that attracts the satellite towards the central body and can be represented as

$$F _ {g r a v} = \frac {\left(G * M _ {s a t} * M _ {e a r t h}\right)}{R _ {o r b i t} ^ {2}}$$

If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting satellite is given by the relationship

$$F _ {n e t} = \frac {\left(M _ {s a t} * v ^ {2}\right)}{R _ {o r b i t}}$$

Since $F_{\text{grav}} = F_{\text{net}}$ , the above expressions for centripetal force and gravitational force can be set equal to each other. Thus,

$$v ^ {2} = \frac {\left(G * M _ {e a r t h}\right)}{R _ {o r b i t}}$$

Taking the square root of each side, leaves the following equation for the velocity of a satellite moving about a central body in circular motion

$$v = \sqrt {\frac {G M _ {e a r t h}}{R _ {e a r t h} + h}}$$

Period of revolution or time period of a satellite

It is the time taken by a satellite to complete one revolution around the Earth. Circumference of a circle of radius $R + h$ is $2\pi (R + h)$ .

Let $T$ be the time period, $v$ the velocity and $h$ the height of the satellite above the Earth's surface.

$$T = \frac {2 \pi (R + h)}{v}$$ $$v = \sqrt {\frac {G M}{R + h}}$$

Substituting in the equation, we get

$$T = \frac {2 \pi (R + h)}{\sqrt {\frac {G M}{R + h}}}$$ $$T = \frac {2 \pi (R + h) \sqrt {R + h}}{\sqrt {G M}} = 2 \pi \sqrt {\frac {(R + h) ^ {3}}{G M}}$$

The gravitational acceleration is

$$g = \frac {G M}{R ^ {2}}$$

Thus,

$$T = 2 \pi \sqrt {\frac {(R + h) ^ {3}}{g R ^ {2}}}$$ $$T = \frac {2 \pi}{R} \sqrt {\frac {(R + h) ^ {3}}{g}}$$

(ii)

Linear Harmonic oscillator.

Mechanical model: mass m on a spring characterized by a spring constant k.

Elastic restoring force $F = -kx$ is balanced according to Newton's second law

$$F = m a$$ $$m \ddot {x} = - k x$$

The equation of motion

$$\ddot {x} + \omega^ {2} x = 0$$

where

$$\omega = \sqrt {\frac {k}{m}}$$

Such system oscillates with amplitude A and angular frequency $\omega$.

Consider solution

$$x = A \sin (\omega t + \varphi)$$

The velocity $\mathbf{v}$ is equal to

$$v = \dot {x} = A \omega \cos (\omega t + \varphi)$$

and thus the kinetic energy

$$K = \frac {1}{2} m v ^ {2} = \frac {1}{2} m A ^ {2} \omega^ {2} \cos^ {2} (\omega t + \varphi) = K _ {0} \cos^ {2} (\omega t + \varphi)$$

where the maximum kinetic energy is equal to

$$K _ {0} = \frac {1}{2} m A ^ {2} \omega^ {2} = \frac {1}{2} k A ^ {2}$$

The potential energy – work done by applied force displacing the system from 0 to x

$$U (x) = \int_ {0} ^ {x} k x d x = \frac {1}{2} k x ^ {2}$$

Substituting $x$

$$U (x) = \frac {1}{2} k A ^ {2} \sin^ {2} (\omega t + \varphi) = U _ {0} \sin^ {2} (\omega t + \varphi)$$

where $\mathsf{U}_0$ is the maximum potential energy (for $x = A$ )

$$U _ {0} = \frac {1}{2} k A ^ {2}$$

The average values over one oscillation period are calculated using the definition

$$\langle f \rangle = \frac {1}{t _ {2} - t _ {1}} \int_ {t _ {1}} ^ {t _ {2}} f (t) d t$$

Thus:

$$\langle U \rangle = \frac {\int_ {0} ^ {T} U d t}{\int_ {0} ^ {T} d t} = \frac {\int_ {0} ^ {T} U _ {0} \sin^ {2} (\omega t + \varphi) d t}{T} = \frac {1}{2} U _ {0} = \frac {1}{2} k A ^ {2}$$

and

$$\langle K \rangle = \frac {\int_ {0} ^ {T} K d t}{\int_ {0} ^ {T} d t} = \frac {1}{2} K _ {0} = \frac {1}{2} k A ^ {2}$$

The sum of the kinetic and potential energies in a simple harmonic oscillator is a constant, i.e., $\mathrm{KE + PE =}$ constant. The energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates.

Thus,

$$\langle E \rangle = \langle K \rangle + \langle U \rangle = \frac {1}{2} k A ^ {2} + \frac {1}{2} k A ^ {2} = k A ^ {2} = m A ^ {2} \omega^ {2}$$

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