Orbital Motion

This note concentrates on the mathematical explanations and derivations presented in the transcript on orbital mechanics. It covers Newton’s law of gravity, the need for horizontal velocity to achieve orbit, derivations of orbital velocity and period, application to a real-world problem, and additional insights regarding trajectory shapes and weightlessness.

Newton’s Law of Gravity and the Need for Horizontal Velocity

Let's start with the question that every launch has to answer. Gravity pulls a spacecraft toward the centre of the Earth all the time, and nothing can switch that pull off. So how does a spacecraft stay up? The answer is speed along the surface, not height above it.

Concept Introduction: The rockets do not simply go “straight up” into space. Instead, to achieve orbit, a spacecraft must build a significant horizontal (tangential) velocity so that while gravity pulls it toward Earth, its path curves away from the surface.
Thought Experiment: Using the example of throwing a rock from a tall mountain, it is shown that with a low horizontal speed the rock follows a parabolic path and eventually hits the ground. Increasing the horizontal speed leads to trajectories that nearly “miss” the ground, eventually resulting in a complete orbit when the horizontal speed is high enough.

We can put numbers on the thought experiment. The curved surface of the Earth drops about 5 m below a straight horizontal line over a distance of 8 km. A dropped object also falls about 4.9 m in its first second. So an object that moves about 8 km sideways in one second falls by about the same amount that the ground curves away below it. The object keeps falling, but it never gets closer to the ground. This speed is the circular orbital velocity at the surface. The Deriving the Orbital Velocity section derives it exactly as v = sqrt(G M / R), which gives about 7.9 km/s.

The direction of the speed matters as much as its size. A rocket climbs almost vertically at first, but only to leave the dense lower atmosphere quickly. It then tilts over, and most of its fuel goes into horizontal speed.

An orbit is continuous free fall : the object falls toward the Earth all the time, and the surface curves away at the same rate.
Horizontal speed decides the orbit, not altitude : a rocket that reaches 400 km of height without about 7.7 km/s of horizontal speed makes only a sub-orbital hop.
The speed needed near the surface is about 7.9 km/s : the value comes from the force balance in the Deriving the Orbital Velocity section.

Trajectory Shapes

The shape of an orbit is determined by the initial velocity of an object relative to the central gravitational force. These shapes—parabola, ellipse, circle, and hyperbola—are mathematical solutions to orbital motion under gravitational influence. Below is a detailed explanation of each type of orbit, how they arise, and their unique characteristics.

Parabolic Trajectories

The word parabola needs a warning before you read the list below. The list uses it for a short throw near the ground. Over such a short path, gravity has almost the same direction and the same strength everywhere, and under that flat-Earth approximation the path is a parabola. On the scale of the whole planet, gravity points to the centre of the Earth. A sub-orbital path is then really a piece of an ellipse that crosses the surface. In orbital mechanics, a parabolic orbit means something else. It is the boundary case at exactly the escape velocity, between the ellipse and the hyperbola.

How They Occur: - When an object is launched with an initial velocity that is insufficient to maintain a stable orbit, it follows a parabolic path. - Gravity pulls the object back to the central body before it completes a full revolution.
Characteristics: - The object rises along a curved path, slows as it reaches its peak, and then accelerates back toward the surface, intersecting with the ground. - The trajectory is symmetrical and governed by the laws of projectile motion.
Examples: - A stone thrown horizontally or upward from a hill. - Sub-orbital rockets that briefly leave the atmosphere but do not achieve orbital speed.

Elliptical Orbits

Elliptical orbits are the most common type of orbit for natural celestial bodies and many artificial satellites. They occur when the object’s initial velocity is sufficient to sustain a closed orbit but does not match the speed needed for a perfect circular path.

Two Key Scenarios:
- Near-Earth Elliptical Orbit:
  - How It Happens: The object’s speed is only slightly above the threshold required to avoid hitting the ground.
  - Key Characteristics: - Speed is highest at perigee due to the stronger gravitational pull. - Speed is lowest at apogee, where gravity’s pull weakens.
  - Practical Examples: Satellites in low Earth orbit (LEO) often follow near-Earth ellipses, with higher speeds near perigee to maintain their trajectory.
- Farther Elliptical Orbit:
  - How It Happens: The object is launched with more energy, resulting in an orbit with a much higher apogee (greater eccentricity).
  - Key Characteristics: - The difference between apogee and perigee becomes more pronounced. - Requires greater horizontal speed than a near-Earth ellipse.
  - Energy Distribution: - At perigee: Kinetic energy is maximized, and potential energy is minimized. - At apogee: Potential energy is maximized, and kinetic energy is minimized.

Conservation of angular momentum gives the exact ratio between the two speeds. At perigee and at apogee the velocity is perpendicular to the radius, so r_p v_p = r_a v_a. The speed at perigee divided by the speed at apogee therefore equals the apogee distance divided by the perigee distance. For example, an ellipse whose apogee distance is twice its perigee distance has a perigee speed twice its apogee speed.

Circular Orbit

The circular orbit needs one exact speed, and it sits between the two kinds of ellipse above. Launch horizontally a little slower, and the launch point becomes the apogee of an ellipse. Launch a little faster, and the launch point becomes the perigee. The next section derives this exact speed.

How It Happens: Achieved when the object’s horizontal velocity is exactly equal to the velocity required to balance gravitational pull at a given altitude.
Key Characteristics: - Speed is constant throughout the orbit. - The orbit’s radius is fixed, and the shape is perfectly symmetrical.
Practical Applications: Common for geostationary satellites, which maintain a fixed position relative to a point on Earth’s surface.

Hyperbolic Trajectories

An object escapes from the planet and never returns once its speed reaches the escape velocity, v_esc = sqrt(2 G M / R). This is sqrt(2), or about 1.41, times the circular speed at the same distance R from the centre. At the surface of the Earth it is about 11.2 km/s. Exactly at v_esc the path is a parabola, and above v_esc the path is a hyperbola.

How It Happens: The object gains sufficient kinetic energy to overcome the gravitational potential energy of the central body completely.
Key Characteristics: - The trajectory does not form a closed loop. - As the object moves away, its velocity gradually decreases but never reaches zero.
Practical Examples: - Spacecraft leaving Earth for missions to other planets. - Comets or asteroids passing near Earth on paths that take them back into deep space.

Visualizing the Transition Between Orbits

The list below orders the cases by launch speed. One number sorts all of them: the specific orbital energy e = v²/2 - G M / R, which is the kinetic energy plus the potential energy per kilogram. A negative e gives a closed orbit, a circle or an ellipse. Zero gives the parabola at the escape velocity, and a positive e gives a hyperbola.

Low Speed: Leads to parabolic paths that intersect with the surface.
Moderate Speed: Results in elliptical orbits, with the eccentricity increasing as speed increases.
Exact Speed for a Circular Orbit: Achieves a perfectly symmetrical orbit with a fixed radius.
High Speed (Above Escape Velocity): Leads to hyperbolic trajectories that escape the gravitational influence of the planet.

Figure 1 draws Newton's mountain to scale. All six paths start horizontally from the same launch point at the top of the mountain, 0.15 Earth radius above the surface. Each path is labelled with its launch speed as a multiple of v_c, the circular speed at the launch height. The paths are computed from the exact conic solution, so the curves are true ellipses, a true parabola and a true hyperbola.

Figure 1. Horizontal launches from one point at six speeds. Below v_c the launch point is the apogee, and above v_c it is the perigee. From 1.41 v_c upward the path no longer closes.

The two red and orange paths fall back : they are pieces of ellipses that cross the surface, with the launch point at the apogee.
The blue circle keeps a constant distance : at exactly v_c the fall and the curvature of the path match all the way round.
The green ellipse is only 15 percent faster than the circle : its far point already lies more than 2 Earth radii from the centre.
The purple parabola and the grey hyperbola do not return : both leave the drawing, and the hyperbola bends less because it carries more energy.

Key Insights

The three points below summarise the section in terms of energy. The derivations that follow switch to forces for the circular case, because a force balance is the shortest route to the speed and the period.

Energy’s Role: The balance between kinetic and potential energy determines the shape and stability of the orbit.
Speed and Shape: The faster the object’s horizontal speed, the less bound the orbit becomes, transitioning from circular to elliptical to hyperbolic.
Applications: Each orbit type serves unique purposes in science and exploration, from launching satellites to sending spacecraft to distant planets.

Deriving the Orbital Velocity

Now we can turn the thought experiment into a formula. The circular orbit is the easiest case to derive, because the speed and the distance to the centre stay constant. So gravity only has to supply the centripetal force, and one force balance gives the speed.

In Circular Orbit

The derivation has three steps. First we write the gravitational force, then the centripetal force that circular motion needs, and finally we set the two equal. Keep in mind that R is measured from the centre of the Earth, not from the surface.

Newton’s Gravitational Force

Gravity is the only force on the orbiting object once the engine stops, so it is the force we start from. Its size falls with the square of the distance between the two centres.

The gravitational force between two bodies is given by

F = (G M m) / R²,

where G is the gravitational constant, M is the mass of the central body (e.g., Earth), m is the mass of the orbiting object, and R is the distance from the center of the planet to the object.

Centripetal Force Requirement

An object that moves on a circle changes its direction all the time, even at a constant speed. That change of direction is an acceleration toward the centre, and a force has to cause it.

For an object in circular motion, the required centripetal acceleration is

a_c = v² / R,

so the centripetal force needed is

F_c = m (v² / R).

Equating Forces

In orbit, gravity is the only force available to supply the centripetal force. So the two expressions above must be equal, and this one equation gives the orbital speed.

Setting the gravitational force equal to the centripetal force gives:

(G M m) / R² = m (v² / R).

Canceling the mass m results in

v² = (G M) / R,

so the orbital velocity is:

v = sqrt(G M / R).

Key Point: The orbital speed depends only on the mass of the planet and the orbital radius; it does not depend on the mass of the orbiting object.

Two checks make the formula concrete. For the Earth, G M = 6.67 x 10^-11 x 5.97 x 10²⁴, which is about 3.98 x 10¹⁴ m³/s². At the surface, R = 6.38 x 10⁶ m, and v comes out at about 7.9 km/s. This is the speed that the thought experiment in the first section asked for.

The escape velocity follows from energy rather than from force. The object escapes when its kinetic energy m v²/2 equals the depth of the gravitational potential well, G M m / R. So v_esc = sqrt(2 G M / R), and the mass m cancels again. At the surface this gives about 11.2 km/s.

A higher orbit is a slower orbit : v goes as 1 / sqrt(R), so a satellite far from the Earth moves more slowly than one close to it.
The mass of the satellite cancels : a small cubesat and a large space station at the same R move at the same speed.
The escape speed is sqrt(2) times the circular speed : this ratio of about 1.41 holds at every distance R.

Deriving the Orbital Period

The speed alone does not tell you how often a satellite passes overhead. For that we need the period, the time for one full orbit. The period follows directly from the speed, because the satellite covers one circumference per orbit at a constant speed.

In Circular Orbit

The steps below only rearrange one expression, T = 2πR / v. Watch the powers of R as you go. One power of R comes from the circumference, and another half power comes from the speed. That is why the result goes as R^3/2.

Definition

The orbital period T is the time it takes to complete one full orbit. For a circular orbit, the distance traveled is the circumference, which is 2πR.

Relation to Velocity

At a constant speed, time equals distance divided by speed. One orbit covers the circumference 2πR, so the period is that distance divided by the orbital speed v.

The period is given by T = (2πR) / v.

Substitution of Orbital Velocity

We now remove v from the expression, so that the period depends only on the orbit radius and the mass of the planet. Each line below changes only one factor of the line before it.

Substituting v = sqrt(G M / R) into the period equation yields:

T = 2πR / sqrt((G M)/R)

this can be rewriten as follows

T = 2πR (1/ sqrt((G M/R)))

this can be rewritten as

T = 2πR (sqrt(1)/ sqrt((G M/R)))

this can be rewritten as

T = 2πR (sqrt(1/(G M/R))

this can be rewritten as

T = 2πR (sqrt(R/(G M))

this can be rewritten as

T = 2πR (R^1/2/sqrt(G M))

which simplifies to:

T = 2π R^3/2 / sqrt(G M).

Interpretation: This formula shows that the orbital period increases with the orbital radius (specifically with the 3/2 power of R) and decreases with the square root of the planet's mass.

Squaring both sides gives the circular-orbit form of Kepler's third law, T² = (4π² / G M) R³. The square of the period is proportional to the cube of the radius. The same law holds for an elliptical orbit when R is replaced by the semi-major axis a.

We can also use the formula in the other direction. A geostationary satellite must circle the Earth once per sidereal day, which is 86,164 s. Solving Kepler's third law for R with G M = 3.98 x 10¹⁴ m³/s² gives R of about 42,150 km from the centre. This is about 35,770 km above the surface, and the orbital speed there is only about 3.07 km/s.

The period grows as R^3/2 : doubling the radius makes one orbit about 2.83 times longer.
The period is divided by sqrt(G M), not multiplied by it : a heavier planet pulls harder, so the satellite moves faster and the period is shorter.
A required period fixes the radius : this is why every geostationary satellite sits at the same altitude.

Application to a Real-World Problem

Let's put real numbers into the two formulas. The example below is a satellite in low Earth orbit, and it shows the one step that is easy to forget. The formulas need the distance from the centre of the Earth, so the altitude must first be added to the radius of the Earth.

Example 1

Low Earth orbit covers altitudes up to about 2,000 km, and 780 km sits well inside that range. The example uses rounded constants, so a calculator with more digits gives slightly different values.

Example Calculation: The transcript provides an example where a satellite orbits 780 km above the Earth’s surface.
Determining Orbital Radius: R = (Earth’s radius) + (altitude). With Earth’s radius ≈ 6.38×10⁶ m and altitude = 780 km = 7.80×10⁵ m, we get: R ≈ 6.38×10⁶ m + 7.80×10⁵ m = 7.16×10⁶ m.
Calculating Orbital Velocity: Using the formula v = sqrt(G M / R) with: G = 6.67×10^-11 N·m²/kg² and M = 5.97×10²⁴ kg, the calculation becomes: v = sqrt((6.67×10^-11 × 5.97×10²⁴) / 7.16×10⁶), resulting in approximately 7.46×10³ m/s (or 7.46 km/s).

The same inputs give the period. With T = 2π R^3/2 / sqrt(G M) and R = 7.16 x 10⁶ m, T comes out at about 6,030 s, or about 100.5 minutes. So the satellite completes about 14.3 orbits per day.

Two more numbers from the same inputs are useful for comparison. The escape velocity at this radius is sqrt(2) x 7.46 km/s, which is about 10.5 km/s. The gravitational acceleration there is G M / R², about 7.77 m/s². That is still about 79 percent of its value at the surface, and the next section uses this number.

Always add the radius of the Earth : using 780 km as R gives a speed of about 22.6 km/s, which is wrong by a factor of three.
A low Earth orbit means about 7.5 km/s and about 100 minutes : these two numbers are a quick check for any result in this altitude range.

Why does an astronaut in orbit feel weightless ?

Gravity at a low Earth orbit is still strong. The example above gives about 7.77 m/s² at 780 km, which is about 79 percent of the value at the surface. So why does an astronaut float inside the spacecraft?

The feeling of weight does not come from gravity itself. It comes from the floor or the seat that pushes back on the body. In orbit, the astronaut and the spacecraft fall toward the Earth with the same acceleration, G M / R². This acceleration does not depend on the mass of the falling object, as the Deriving the Orbital Velocity section showed. So the floor falls away exactly as fast as the astronaut falls toward it. The floor then pushes with zero force, and the astronaut feels no weight.

This is the thought experiment of the first section again. The orbit is a continuous free fall, and everything inside the spacecraft takes part in the same fall. A falling elevator would give the same feeling for a short moment. Aircraft for zero-g training fly parabolic arcs to produce the same state for about 20 to 25 seconds at a time.

The word microgravity is more accurate than zero gravity. Small effects remain. The thin upper atmosphere drags on the spacecraft, and gravity pulls slightly harder on the parts of the spacecraft nearer the Earth. So loose objects inside drift slowly rather than float perfectly still.

Weightless does not mean without gravity : at 780 km, gravity still gives an acceleration of about 7.77 m/s².
Weight is the contact force, not the pull of gravity : a scale inside an orbiting spacecraft reads zero, because the scale falls together with the person on it.
The same cancellation of mass explains both results : the orbital speed does not depend on m, and for the same reason every object in the spacecraft falls together.

YouTube

Intro to Orbital Motion & Orbital Mechanics - Math and Science (2025)