# A satellite of mass m is in a circular earth orbit of radius r. (a) Give an algebraic expression...

## Question:

A satellite of mass m is in a circular earth orbit of radius r.

(a) Give an algebraic expression for the satellite's mechanical energy in terms of the mass of the Earth (M), the gravitational constant (G), m, and r.

(Gravitational potential energy is a negative quantity under these circumstances.)

(b) Calculate the satellite's energy and speed if m = {eq}1.70 \times 10^4 {/eq} kg and r = {eq}1.00 \times 10^7 {/eq} m.

## Circular Motion:

Satellites orbit the earth using the principle of centripetal motion where the acceleration is directed towards an equilibrium point, in this case at the center of the earth, and the radius of orbit of the satellite is maintained by the sufficiently high centripetal speed. The mechanical energy of a satellite equals the sum of its gravitational potential energy and kinetic energy.

## Answer and Explanation:

#### Part a)

For a satellite of mass *m* in the earth's orbit of radius *r* at a speed *v* where the mass of the Earth (M), and the gravitational constant (G), the mechanical energy is given by the equation:

{eq}\begin{align*} \text{Mechanical Energy} &= \text{P.E}+ \text{K.E}\\ &= -\frac{GMm}{r} + \frac{1}{2}mv^2\\ &= -\frac{GMm}{r} + \frac{1}{2}m(\frac{GM}{r})\ \ \ \text{because}\ \ \ \ v^2 = \frac{GM}{r}\\ &\color{blue}{= \boxed{-\frac{GMm}{2r}}} \end{align*} {/eq}

#### Part b)

We have the following:

- The mass of the earth is {eq}M = 5.97\ \times 10^{24}\ \text{kg} {/eq}

- The mass of the satellite is {eq}m = 1.70 \times 10^4\ \text{kg} {/eq}

- The radius of the orbit is {eq}r = 1.00 \times 10^7\ \text{m} {/eq}

- the value of {eq}G = 6.7\times 10^{-11}\ \text{ N?m}^2\text{kg}^2 {/eq}

Using a suitable equation,

{eq}\begin{align*} \text{Mechanical Energy} &= \frac{GMm}{2r}\\ &= -\frac{ 6.67\times 10^{-11}\ \text{ Nm}^2\text{kg}^2(5.97\ \times 10^{24}\ \text{kg})( 1.70 \times 10^4\ \text{kg})}{2(1.00 \times 10^7\ \text{m})}\\ &\color{blue}{\approx \boxed{3.38\ \times 10^{11}\ \text{J}}} \end{align*} {/eq}

Using a suitable equation,

{eq}\begin{align*} v^2 &= \frac{GM}{r}\\ v &= \sqrt{\frac{GM}{r}}\\ &= \sqrt{\frac{6.67\times 10^{-11}\ \text{ Nm}^2\text{kg}^2(5.97\ \times 10^{24}\ \text{kg})}{1.00 \times 10^7\ \text{m}}}\\ &\color{blue}{\approx \boxed{6\ 310\ \text{m/s}}} \end{align*} {/eq}