Orbital Motion — Revision Notes

⚡ 30-Second Revision

Orbital Velocity: — $v_o = \sqrt{\frac{GM}{r}}$ (independent of satellite mass $m$)
Time Period: — $T = \frac{2\pi r}{v_o} = 2\pi \sqrt{\frac{r^3}{GM}}$ (Kepler's 3rd Law: $T^2 \propto r^3$)
Kinetic Energy: — $K = \frac{1}{2}mv_o^2 = \frac{GMm}{2r}$
Potential Energy: — $U = -\frac{GMm}{r}$
Total Energy: — $E = K + U = -\frac{GMm}{2r}$
Binding Energy: — $-E = \frac{GMm}{2r}$
Relationship: — $K = -E$, $U = 2E$
Orbital Radius: — $r = R_E + h$ (where $R_E$ is Earth's radius, $h$ is altitude)
Geostationary Satellite: — $T = 24,\text{hours}$, orbits in equatorial plane, $h \approx 35,786,\text{km}$.
Weightlessness: — Due to continuous freefall, not absence of gravity.
Escape Velocity: — $v_e = \sqrt{2} v_o = \sqrt{\frac{2GM}{r}}$

2-Minute Revision

Orbital motion is governed by gravity providing the necessary centripetal force. The orbital velocity ($v_o$) of a satellite of mass $m$ around a central body of mass $M$ at radius $r$ is $v_o = sqrt{GM/r}$, crucially independent of $m$.

The time period ($T$) for one revolution is $T = 2pi sqrt{r^3/GM}$, which is Kepler's Third Law ($T^2 propto r^3$). An orbiting satellite possesses kinetic energy ($K = GMm/2r$) and gravitational potential energy ($U = -GMm/r$).

Its total mechanical energy ($E = -GMm/2r$) is negative, indicating it's gravitationally bound. Key relationships are $K = -E$ and $U = 2E$. Geostationary satellites are special: they have a 24-hour period, orbit in the equatorial plane at a specific altitude ($35,786,\text{km}$ above Earth's surface), and appear stationary.

The 'weightlessness' experienced in orbit is due to continuous freefall, not zero gravity. Remember that orbital radius $r$ is measured from the center of the central body ($R_E + h$).

5-Minute Revision

Orbital motion is a direct application of Newton's Law of Universal Gravitation, where the gravitational force ($F_g = GMm/r^2$) acts as the centripetal force ($F_c = mv_o^2/r$) to keep an object in orbit.

Equating these yields the orbital velocity $v_o = sqrt{GM/r}$. Notice that $v_o$ is independent of the orbiting mass $m$, a common conceptual trap. Higher orbits mean lower orbital velocities. The time period $T$ for one revolution is $T = 2pi r / v_o$, which simplifies to $T = 2pi sqrt{r^3/GM}$.

This is Kepler's Third Law, $T^2 propto r^3$, essential for ratio problems. Remember to use $r = R_E + h$ for orbital radius, where $R_E$ is the central body's radius and $h$ is the altitude.

The energy considerations are vital: Kinetic Energy $K = \frac{1}{2}mv_o^2 = \frac{GMm}{2r}$. Gravitational Potential Energy $U = -\frac{GMm}{r}$. The total mechanical energy $E = K + U = -\frac{GMm}{2r}$.

The negative sign of $E$ signifies that the satellite is bound to the central body. The binding energy, or energy required to escape, is $-E = \frac{GMm}{2r}$. Crucial relationships are $K = -E$ and $U = 2E$.

Geostationary satellites are a specific application: they have a 24-hour period, orbit in the equatorial plane at an altitude of approximately $35,786,\text{km}$ above Earth's surface, and appear stationary.

The sensation of weightlessness in orbit is due to continuous freefall, where the satellite and its contents are constantly accelerating towards Earth together, not because gravity is absent. Finally, distinguish orbital velocity from escape velocity ($v_e = sqrt{2} v_o$).

Prelims Revision Notes

1
Orbital Velocity ($v_o$): — $v_o = \sqrt{\frac{GM}{r}}$. It is independent of the mass of the orbiting satellite ($m$). $r$ is the distance from the center of the central body ($r = R_E + h$).
2
Time Period ($T$): — $T = \frac{2\pi r}{v_o} = 2\pi \sqrt{\frac{r^3}{GM}}$. This is Kepler's Third Law ($T^2 \propto r^3$). Use this for ratio problems.
3
Relationship between $GM$ and $g$: — $g = \frac{GM}{R_E^2} \implies GM = gR_E^2$. Substitute this into formulas when $g$ is given.
4
Energy of a Satellite:

* Kinetic Energy ($K$): $K = \frac{1}{2}mv_o^2 = \frac{GMm}{2r}$. Always positive. * Potential Energy ($U$): $U = -\frac{GMm}{r}$. Always negative. * Total Energy ($E$): $E = K + U = -\frac{GMm}{2r}$. Always negative for bound orbits. * Binding Energy: Energy required to escape orbit, equal to $-E = \frac{GMm}{2r}$.

1
Energy Relationships: — $K = -E$ and $U = 2E$. These are frequently tested.
2
Geostationary Satellites:

* Orbital Period: $T = 24,\text{hours}$ (matches Earth's rotation). * Orbital Plane: Equatorial plane. * Altitude: Approximately $35,786,\text{km}$ above Earth's surface ($r \approx 42,164,\text{km}$ from center). * Appears stationary from Earth's surface.

1
Weightlessness in Orbit: — It's a state of continuous freefall, not absence of gravity. Astronauts and objects inside a spacecraft fall together.
2
Escape Velocity ($v_e$): — $v_e = \sqrt{\frac{2GM}{r}} = \sqrt{2} v_o$. Minimum velocity to escape gravitational field. Total energy is zero at escape velocity.
3
Key Proportionalities: — $v_o \propto r^{-1/2}$, $T \propto r^{3/2}$, $K \propto r^{-1}$, $U \propto r^{-1}$, $E \propto r^{-1}$. Be quick with these for ratio questions.

Vyyuha Quick Recall

VET KUTE

Velocity: $v_o = \sqrt{GM/r}$
Escape: $v_e = \sqrt{2} v_o$
Time Period: $T^2 \propto r^3$
Kinetic Energy: $K = -E$
U — Potential Energy: $U = 2E$
Total Energy: $E = -GMm/2r$ (always negative)
Equatorial (for Geostationary)