Motion of Orbiting Satellites · AP® Physics 1 AP Physics 1

Why a satellite stays up: orbital speed

Picture a cannon on a very tall mountain. A slow cannonball curves down and lands nearby. A faster one lands farther away. If the ball is fast enough, Earth's surface curves away as fast as the ball drops, and the ball never lands. That ball is now a satellite.

Only one force acts on the satellite: gravity, $F_g$, pointing toward Earth's center. (The "centripetal force" isn't a separate force. It's the name for the net force toward the center, and here it equals $F_g$.) For circular motion, Newton's second law in the radial direction says $F_g = m a_c$:

$$\frac{G M m}{r^2} = m\,\frac{v^2}{r}$$

The satellite's mass $m$ appears on both sides and cancels. Solving for $v$:

Why it makes sense: a larger $r$ means weaker gravity, so less speed is needed to "keep missing" the planet. The mass of the satellite doesn't matter, because a heavier satellite feels proportionally more gravity but also has proportionally more inertia.

Speed and period of a space-station orbit

A station orbits Earth at an altitude of 400 km. Find its speed. Use $R_E = 6.37\times10^{6}\ \text{m}$ and $M_E = 5.97\times10^{24}\ \text{kg}$.

Find $r$ from Earth's center. $r = R_E + h = 6.37\times10^{6} + 4.0\times10^{5} = 6.77\times10^{6}\ \text{m}$.

Substitute into $v = \sqrt{GM/r}$. $GM_E = (6.67\times10^{-11})(5.97\times10^{24}) = 3.98\times10^{14}\ \text{N·m}^2/\text{kg}$.

$$v = \sqrt{\frac{3.98\times10^{14}}{6.77\times10^{6}}} = \sqrt{5.88\times10^{7}} \approx 7.67\times10^{3}\ \text{m/s}$$

Your turn: Use $T = 2\pi r / v$ to find the time for one lap, in minutes.

$T = \dfrac{2\pi (6.77\times10^{6})}{7.67\times10^{3}} \approx 5.55\times10^{3}\ \text{s} \approx 92.5\ \text{min}$. The station circles Earth about 15.6 times a day. Equivalently, the angular speed is $\omega = v/r = 2\pi/T \approx 1.13\times10^{-3}\ \text{rad/s}$, and $a_c = \omega^2 r = v^2/r$ gives the same orbit equation.

Practice★☆☆

A satellite moves in a circular orbit of radius $7.0\times10^{6}\ \text{m}$ measured from Earth's center. Use $M_E = 5.97\times10^{24}\ \text{kg}$ and…

A satellite moves in a circular orbit of radius $7.0\times10^{6}\ \text{m}$ measured from Earth's center. Use $M_E = 5.97\times10^{24}\ \text{kg}$ and $G = 6.67\times10^{-11}\ \text{N·m}^2/\text{kg}^2$. Calculate the satellite's orbital speed. Does the answer depend on the satellite's mass?

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Practice★★☆

A probe orbits Mars in a circular orbit $4.00\times10^{5}\ \text{m}$ above the surface. Mars has mass $6.42\times10^{23}\ \text{kg}$ and radius…

A probe orbits Mars in a circular orbit $4.00\times10^{5}\ \text{m}$ above the surface. Mars has mass $6.42\times10^{23}\ \text{kg}$ and radius $3.39\times10^{6}\ \text{m}$. Use $G = 6.67\times10^{-11}\ \text{N·m}^2/\text{kg}^2$. (a) Calculate the orbital radius measured from Mars's center. (b) Calculate the probe's orbital speed. (c) Calculate the orbital period in minutes.

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Period and Kepler's third law

One lap is a distance of $2\pi r$ at speed $v$, so the period is $T = 2\pi r / v$. Substitute $v=\sqrt{GM/r}$:

$$T = 2\pi\sqrt{\frac{r^3}{GM}} \quad\Longrightarrow\quad T^2 = \frac{4\pi^2}{GM}\,r^3$$

The farther out a satellite is, the longer its lap and the slower it moves. Both effects make the period grow quickly with $r$.

You can also solve for the radius when you know the period, which is how engineers choose an orbit:

$$r = \left(\frac{G M T^2}{4\pi^2}\right)^{1/3}$$

A geostationary satellite has a period of one sidereal day (about $8.64\times10^{4}\ \text{s}$), so it stays above the same spot on Earth. The orbit must be above the equator. Gravity always points to Earth's center, so the orbital plane always contains Earth's center. Only an orbit above the equator can stay over a fixed point on the surface as Earth turns.

Practice★☆☆

The Moon orbits Earth in a nearly circular path of radius $3.84\times10^{8}\ \text{m}$ with a period of $27.3$ days. Convert the period to seconds and calculate the…

The Moon orbits Earth in a nearly circular path of radius $3.84\times10^{8}\ \text{m}$ with a period of $27.3$ days. Convert the period to seconds and calculate the Moon's average orbital speed. Give your answer in m/s.

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Practice★★☆

A communications satellite is placed in a geostationary orbit, so it stays above the same point on the equator. Its period is $8.64\times10^{4}\ \text{s}$. Use…

A communications satellite is placed in a geostationary orbit, so it stays above the same point on the equator. Its period is $8.64\times10^{4}\ \text{s}$. Use $M_E = 5.97\times10^{24}\ \text{kg}$, $R_E = 6.37\times10^{6}\ \text{m}$, and $G = 6.67\times10^{-11}\ \text{N·m}^2/\text{kg}^2$. (a) Derive an expression for the orbital radius in terms of $G$, $M_E$, and the period. (b) Calculate the orbital radius and the altitude above Earth's surface. (c) Explain why a geostationary orbit must lie above the equator.

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Comparing orbits with ratios

Many questions don't need numbers. They ask what happens to $v$, $T$, or $F_g$ when $r$ changes. Write each quantity as a power of $r$:

Quantity	Equation	Depends on $r$ as
Orbital speed	$v=\sqrt{GM/r}$	$v\propto r^{-1/2}$
Period	$T = 2\pi\sqrt{r^3/GM}$	$T\propto r^{3/2}$
Gravitational force	$F_g = GMm/r^2$	$F_g\propto r^{-2}$

Practice★☆☆

A satellite in a circular orbit around Earth is boosted to a new circular orbit of larger radius. For each quantity, state whether it increases, decreases, or stays the…

A satellite in a circular orbit around Earth is boosted to a new circular orbit of larger radius. For each quantity, state whether it increases, decreases, or stays the same, and justify your choice briefly: (a) orbital speed, (b) orbital period, (c) gravitational force on the satellite.

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Practice★★☆

Satellite A and satellite B have the same mass and both orbit Earth in circular orbits. Satellite A has orbital radius $r$ and satellite B has orbital radius $2r$. Find…

Satellite A and satellite B have the same mass and both orbit Earth in circular orbits. Satellite A has orbital radius $r$ and satellite B has orbital radius $2r$. Find the ratio of each quantity, A to B: (a) orbital speed, (b) orbital period, (c) kinetic energy. Show how each ratio follows from the orbit equations.

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Energy of an orbit

Far from the surface we can't use $\Delta U_g = mg\Delta y$, because $g$ isn't constant. For the Earth–satellite system the potential energy is

$$U = -\frac{G M m}{r}$$

(zero at infinite separation, so it's negative everywhere in an orbit). The kinetic energy comes from $v^2 = GM/r$:

$$K = \tfrac12 m v^2 = \frac{G M m}{2r}$$

Add them:

The total energy is negative: the satellite is bound. A larger orbit has a total energy closer to zero (less negative), so to move out, the engines must add energy.

Raising a satellite

A 500 kg satellite moves from $r_1 = 8.0\times10^{6}\ \text{m}$ to $r_2 = 1.6\times10^{7}\ \text{m}$. Find $\Delta K$, $\Delta U$, and the minimum work by the engines.

Compute $GMm = (3.98\times10^{14})(500) = 1.99\times10^{17}\ \text{J·m}$.

Kinetic energy. $K_1 = \dfrac{1.99\times10^{17}}{2(8.0\times10^{6})} = 1.24\times10^{10}\ \text{J}$ and $K_2 = \dfrac{1.99\times10^{17}}{2(1.6\times10^{7})} = 6.2\times10^{9}\ \text{J}$. So $\Delta K = -6.2\times10^{9}\ \text{J}$. The satellite loses kinetic energy.

Potential energy. $U_1 = -2.49\times10^{10}\ \text{J}$ and $U_2 = -1.24\times10^{10}\ \text{J}$. So $\Delta U = +1.24\times10^{10}\ \text{J}$.

Total. $\Delta E = \Delta K + \Delta U = +6.2\times10^{9}\ \text{J}$. This is the minimum work the engines must do (if friction-free in vacuum, $W_{engines} = \Delta E$).

Your turn: Explain why the speed dropped even though $W_{engines}>0$.

The engines' work raised $E$ by $\Delta E = +6.2\times10^{9}\ \text{J}$, but $\Delta U = +1.24\times10^{10}\ \text{J}$ is twice that. The extra energy needed for $U$ comes at the expense of $K$ ($\Delta K = -6.2\times10^{9}\ \text{J}$). Climbing away from Earth costs kinetic energy, and the larger orbit needs less speed.

Practice★★★

A satellite of mass $1{,}200\ \text{kg}$ is moved from a circular orbit of radius $7.0\times10^{6}\ \text{m}$ to a circular orbit of radius $1.4\times10^{7}\ \text{m}$…

A satellite of mass $1{,}200\ \text{kg}$ is moved from a circular orbit of radius $7.0\times10^{6}\ \text{m}$ to a circular orbit of radius $1.4\times10^{7}\ \text{m}$ around Earth. Use $M_E = 5.97\times10^{24}\ \text{kg}$, $G = 6.67\times10^{-11}\ \text{N·m}^2/\text{kg}^2$, and $U = -GM_Em/r$. (a) Calculate the change in kinetic energy, in gravitational potential energy, and in total mechanical energy. (b) Determine the minimum net work the engines must do. (c) Explain why the satellite's speed decreases even though the engines did positive work on it.

Solve

Elliptical orbits

Most orbits aren't circles. In an ellipse the satellite is closest to Earth at perigee and farthest at apogee. At both of those points the velocity is perpendicular to the line to Earth's center.

Two conservation laws determine the speeds:

Angular momentum. Gravity points toward Earth's center, so it exerts zero torque about Earth's center ($\vec r \parallel \vec F$). Therefore $L$ is constant. At perigee and apogee $r_\perp = r$, so

$$m v_p r_p = m v_a r_a \quad\Longrightarrow\quad v_p r_p = v_a r_a$$

The satellite moves faster when it's closer. This is Kepler's second law.

Energy. Only gravity does work, so $K+U$ is constant:

$$\tfrac12 m v_p^2 - \frac{GMm}{r_p} = \tfrac12 m v_a^2 - \frac{GMm}{r_a}$$

Perigee and apogee speeds

A satellite has $r_p = 8.0\times10^{6}\ \text{m}$ and $r_a = 1.6\times10^{7}\ \text{m}$. Find $v_p$ and $v_a$.

Angular momentum: $v_a = v_p\,\dfrac{r_p}{r_a} = \dfrac{v_p}{2}$.

Energy (the mass cancels): $\tfrac12 v_p^2 - \dfrac{GM}{r_p} = \tfrac12\left(\dfrac{v_p}{2}\right)^2 - \dfrac{GM}{r_a}$.

Collect the $v_p$ terms: $\tfrac12 v_p^2\left(1-\tfrac14\right) = GM\left(\dfrac{1}{r_p}-\dfrac{1}{r_a}\right)$, so $\tfrac38 v_p^2 = \dfrac{3.98\times10^{14}}{1.6\times10^{7}} = 2.49\times10^{7}$.

$v_p^2 = 6.64\times10^{7}$, so $v_p \approx 8.15\times10^{3}\ \text{m/s}$ and $v_a \approx 4.07\times10^{3}\ \text{m/s}$.

Your turn: Compare $v_p$ with the circular-orbit speed at $r_p$.

$v_{circ}=\sqrt{GM/r_p} = \sqrt{3.98\times10^{14}/8.0\times10^{6}} \approx 7.05\times10^{3}\ \text{m/s}$. Since $v_p > v_{circ}$, the satellite is moving too fast to stay at that radius, so it climbs outward to the apogee. At apogee $v_a$ is less than the circular speed there ($\approx 4.99\times10^{3}\ \text{m/s}$), so it falls back inward.

Practice★★★

A satellite moves in an elliptical orbit around Earth. Its closest point (perigee) is at distance $7.0\times10^{6}\ \text{m}$ and its farthest point (apogee) is at…

A satellite moves in an elliptical orbit around Earth. Its closest point (perigee) is at distance $7.0\times10^{6}\ \text{m}$ and its farthest point (apogee) is at distance $2.1\times10^{7}\ \text{m}$, both measured from Earth's center. Use $M_E = 5.97\times10^{24}\ \text{kg}$ and $G = 6.67\times10^{-11}\ \text{N·m}^2/\text{kg}^2$. (a) Use conservation of angular momentum to relate the speeds at perigee and apogee. (b) Combine this with conservation of energy to find both speeds. (c) Compare the perigee speed with the speed of a circular orbit at the same radius, and explain what the comparison tells you about the orbit's shape.

Solve

Summary

Let's practice! 8 exercises in this section

★☆☆ A satellite moves in a circular orbit of radius $7.0\times10^{6}\ \text{m}$ measured from…★☆☆ A satellite in a circular orbit around Earth is boosted to a new circular orbit of larger…★☆☆ The Moon orbits Earth in a nearly circular path of radius $3.84\times10^{8}\ \text{m}$…