Satellite Orbits and Energy

For simplicity, this chapter only covers orbits along a circular path.

Deriving Orbital Velocity

Consider a satellite of mass $m$ orbiting Earth (mass $M_E$ ) in a circular path at a distance $r$ from Earth's center. The only force acting on the satellite is gravity, which acts as the required centripetal force holding the satellite in orbit:

\frac{GmM_E}{r^2} = ma_c

Recall that centripetal acceleration is defined as $a_c = \frac{v^2}{r}$ . Substituting this expression yields:

\frac{GmM_E}{r^2} = m \frac{v^2}{r}

Canceling the satellite's mass $m$ and multiplying both sides by $r$ isolates the squared velocity:

v^2 = \frac{GM_E}{r}

Taking the square root gives the orbital speed:

v_{\text{orbit}} = \sqrt{\frac{GM_E}{r}}

(1)

Applying Orbital Velocity

Using the expression for orbital velocity, we can determine the kinetic energy ( $K$ ) of a satellite in a circular orbit:

K = \frac{1}{2}m v^2 = \frac{1}{2} m \left(\frac{GM_E}{r}\right) = \frac{GM_E m}{2r}

(2)

We can also apply orbital velocity to find the period of circular orbit. If we say the length of the circular orbit is $2 \pi r$ and lasts $T$ seconds, we can derive $T$ with the following:

v_{\text{orbit}} = \frac{2 \pi r}{T}

\sqrt{\frac{GM_E}{r}} = \frac{2 \pi r}{T}

Inverting both sides yields:

\sqrt{\frac{r}{GM_E}} = \frac{T}{2 \pi r}

T = 2 \pi r \sqrt{\frac{r}{GM_E}}

We can further simplify by bringing $r$ inside of the radical:

T = 2 \pi \sqrt{r^2} \sqrt{\frac{r}{GM_E}}

T = 2 \pi \sqrt{\frac{r^3}{GM_E}}

(3)

Total Mechanical Energy in Circular Orbits

Now that we have formulas for kinetic and potential energy for objects in orbit we can find the total mechanical energy:

Recall that the total mechanical energy for an object is the sum of its kinetic and potential energies:

E_{\text{mech}} = K + U

We can apply known $K$ from Equation 2 with our known $U$ from the prior chapter to get $E_{\text{mech}}$ :

E_{\text{mech}} = \frac{GM_Em}{2r} - \frac{GM_Em}{r}

Simplifying, we get the following:

E_{\text{mech}} = \frac{GM_Em}{2r} - \frac{2GM_Em}{2r}

E_{\text{mech}} = \frac{GM_Em - 2GM_Em}{2r}

E_{\text{mech}} = -\frac{GM_Em}{2r}

(4)

Notice that for a circular orbit, $E_{\text{mech}} = -K = \frac{1}{2}U$ . Because total energy is negative ( $E_{\text{mech}} < 0$ ), the satellite is in a bound orbit.

Energy Required to Reach Orbit

Launching an object into orbit requires supplying energy for two separate things:

Gravitational Potential Energy: Raising the object from Earth's surface ( $r = R_E$ ) to orbital altitude ( $r = R_E + h$ ).
Kinetic Energy: Accelerating the object up to orbital speed ( $v_{\text{orbit}}$ ).

Because the object is at rest on the surface before launch, its initial energy is purely potential:

E_{\text{initial}} = -\frac{GM_E m}{R_E}

Once in orbit, its total energy is:

E_{\text{final}} = -\frac{GM_E m}{2r}

The minimum work required to launch the satellite into orbit is:

W = E_{\text{final}} - E_{\text{initial}} = -\frac{GM_E m}{2r} + \frac{GM_E m}{R_E} = GM_E m \left( \frac{1}{R_E} - \frac{1}{2r} \right)

(5)

Example 1

The International Space Station (ISS) orbits at an altitude of approximately $h = 400 \text{ km}$ above Earth's surface. Given $M_E = 5.97 \times 10^{24} \text{ kg}$ and $R_E = 6.37 \times 10^6 \text{ m}$ , calculate the orbital speed ( $v_{\text{orbit}}$ ) and the orbital period ( $T$ ) of the ISS.

Hint Click to unblur

The radius of the orbit $r$ is measured from the center of Earth: $r = R_E + h$ . Use $v_{\text{orbit}} = \sqrt{\frac{GM_E}{r}}$ and $T = 2\pi \sqrt{\frac{r^3}{GM_E}}$ .

Solution Click to unblur

First, convert altitude to meters and calculate the total orbital radius from Earth's center:

r = R_E + h = 6.37 \times 10^6\text{ m} + 0.40 \times 10^6\text{ m} = 6.77 \times 10^6\text{ m}

To find orbital velocity, substitute $r$ and Earth's mass into Equation 1:

v_{\text{orbit}} = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{(6.674 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2)(5.97 \times 10^{24}\text{ kg})}{6.77 \times 10^6\text{ m}}} \approx 7.67 \times 10^3\text{ m/s}

To find the period, substitute $r$ into Equation 3:

T = 2\pi \sqrt{\frac{r^3}{GM_E}} = 2\pi \sqrt{\frac{(6.77 \times 10^6\text{ m})^3}{(6.674 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2)(5.97 \times 10^{24}\text{ kg})}} \approx 5550\text{ s}

Converting the period into minutes:

T = \frac{5550\text{ s}}{60\text{ s/min}} \approx 92.5\text{ minutes}

Example 2

A $1000\text{ kg}$ satellite is launched from rest on Earth's surface into a circular orbit at an altitude of $h = 400\text{ km}$ ( $r = 6.77 \times 10^6\text{ m}$ ). How much work must be done to place this payload into orbit?

Hint Click to unblur

The required work equals the total change in mechanical energy: $W = E_{\text{final}} - E_{\text{initial}} = -\frac{GM_E m}{2r} - \left(-\frac{GM_E m}{R_E}\right)$ .

Solution Click to unblur

Using the work-energy relation for orbital insertion (Equation 5):

W = GM_E m \left( \frac{1}{R_E} - \frac{1}{2r} \right)

First, calculate the common product $GM_E m$ :

GM_E m = (6.674 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2)(5.97 \times 10^{24}\text{ kg})(1000\text{ kg}) \approx 3.984 \times 10^{17}\text{ N}\cdot\text{m}^2

Now substitute $R_E = 6.37 \times 10^6\text{ m}$ and $r = 6.77 \times 10^6\text{ m}$ :

W = (3.984 \times 10^{17}\text{ N}\cdot\text{m}^2) \left( \frac{1}{6.37 \times 10^6\text{ m}} - \frac{1}{2(6.77 \times 10^6\text{ m})} \right)

W = (3.984 \times 10^{17}\text{ N}\cdot\text{m}^2) \left( 1.570 \times 10^{-7}\text{ m}^{-1} - 0.739 \times 10^{-7}\text{ m}^{-1} \right)

W \approx 3.31 \times 10^{10}\text{ J}