NDA Physics · Gravitation

Orbits, Kepler and Escape

A planet's orbital period and size obey Kepler's third law T² ∝ a³; a satellite stays up on the gravitational force alone (needing no fuel), and escaping a planet's pull requires a launch speed v_e = √(2GM/R) = √(2gR).

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Why this matters

Four PYQs, two of which are the bread-and-butter Kepler-ratio calculations (period ratio from orbit radii, and semi-major axis from a long planetary year). One item asks what actually keeps a satellite in orbit (gravity, no fuel), and a HARD escape-speed item tests how v_e scales when a planet's radius and density change together. Hold T² ∝ a³ and the escape-speed formula and these are reliable marks.

Concept 1 of 4

Kepler's third law — T² ∝ a³

Intuition

The farther a planet orbits, the longer its year — but not in proportion. Kepler found the square of the orbital period grows as the cube of the orbit size. So a planet four times farther out takes 4^(3/2) = 8 times as long to go round.

Definition

Kepler's third law: for bodies orbiting the same central mass, the square of the orbital period is proportional to the cube of the orbit's semi-major axis (for a circular orbit, its radius):

T^2 \propto a^3, \qquad \dfrac{T_1^2}{T_2^2} = \left(\dfrac{a_1}{a_2}\right)^3.

Equivalently

T \propto a^{3/2}

. Take ratios — the constant of proportionality cancels and you never need its numerical value.

Kepler's third law

T^2 \propto a^3 \qquad\Longleftrightarrow\qquad \dfrac{T_1}{T_2} = \left(\dfrac{a_1}{a_2}\right)^{3/2}

Torbital period (the 'year')
asemi-major axis (orbit radius for a circle)

The Sun sits at one focus of the elliptical orbit. The orbit's size is its semi-major axis a, and the square of the period grows as the cube of a: T² ∝ a³.

Worked example

A planet orbits the Sun at four times the Earth's orbital radius. How many Earth-years long is its year?

Kepler: $T \propto a^{3/2}$ , so $\dfrac{T_p}{T_E} = \left(\dfrac{4}{1}\right)^{3/2}$ .
$4^{3/2} = (4^{1/2})^3 = 2^3 = 8$ .
So the planet's year is 8 Earth-years.

Answer:8 Earth-years.

Practice this conceptself-check · 4 quick reps

Try it yourself

One year on a planet is 8 times as long as one Earth-year. How does its orbital semi-major axis compare with Earth's?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Two orbits with radii R and 4R. Ratio of periods T₁/T₂?
2.
Orbit radius made 9 times larger. Period grows by?
3.
Kepler's third law relates which two quantities?
4.
If period quadruples, the orbit radius grows by?

From the bank · past-year question

Example 1GravitationMODERATE

Two planets orbit the Sun in circular orbits, with their radius of orbit as

R_1 = R

and

R_2 = 4R

. Ratio of their periods

(T_1/T_2)

around the Sun will be

[Q128 · Apr · 2020]

It's T² ∝ a³, not T ∝ a

Don't read Kepler's law as 'period proportional to radius'. The square of the period goes as the cube of the radius. A 4× larger orbit gives an 8× longer period (4^(3/2)), not a 4× one.

Drill 1 more on kepler's third law — t² ∝ a³

Concept 2 of 4

Orbital velocity — v_o = √(GM/R)

Intuition

For a satellite in a circular orbit, gravity supplies exactly the centripetal force needed to bend its path into a circle. Setting the gravitational pull equal to mv²/R and solving for the speed gives the one orbital speed that keeps the satellite at that radius — independent of the satellite's own mass.

Definition

For a circular orbit of radius $R$ around a mass $M$ , equating gravity to the centripetal requirement, $\dfrac{GMm}{R^2} = \dfrac{mv_o^2}{R}$ , gives the orbital velocity

v_o = \sqrt{\dfrac{GM}{R}} = \sqrt{gR}\ (\text{at the surface}).

It does not depend on the satellite's mass. A lower orbit (smaller

R

) requires a faster orbital speed.

Orbital velocity

v_o = \sqrt{\dfrac{GM}{R}}

v_ocircular orbital speed at radius R
Mmass of the central body
Rorbit radius (from the centre)

Worked example

Show that the orbital speed of a satellite just above a planet's surface can be written as √(gR), where g is the surface gravity and R the planet's radius.

Orbital speed: $v_o = \sqrt{\dfrac{GM}{R}}$ .
Surface gravity gives $g = \dfrac{GM}{R^2}$ , so $GM = gR^2$ .
Substitute: $v_o = \sqrt{\dfrac{gR^2}{R}} = \sqrt{gR}$ .

Answer:

v_o = \sqrt{gR}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Does the orbital speed depend on the satellite's mass?
2.
A lower orbit needs a faster or slower orbital speed?
3.
Orbital speed at the surface in terms of g and R?
4.
Orbital velocity formula?

Orbital speed is set by the orbit, not the satellite

The satellite's own mass cancels out of v_o = √(GM/R). A heavy and a light satellite at the same radius orbit at exactly the same speed; only the radius (and the planet's mass) sets it.

Concept 3 of 4

Escape velocity — v_e = √(2GM/R) and how it scales

Intuition

Escape velocity is the minimum launch speed that lets a projectile leave a planet for good, never to fall back. It is √2 times the orbital speed: v_e = √(2GM/R) = √(2gR). When a problem changes a planet's radius and density together, rewrite v_e in terms of density to see how it scales — the powers can cancel in surprising ways.

Definition

The escape velocity from the surface of a planet of mass $M$ and radius $R$ is

v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}.

Writing

M = \rho \cdot \tfrac{4}{3}\pi R^3

gives

v_e = R\sqrt{\tfrac{8}{3}\pi G \rho}

, i.e. **

v_e \propto R\sqrt{\rho}

**. So the escape speed scales with the radius times the square root of the density.

Escape velocity and its density scaling

v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}; \qquad v_e \propto R\sqrt{\rho}

v_eescape velocity from the surface
M, Rplanet's mass and radius
\rhoplanet's mean density

Worked example

Earth's escape speed is about 11.2 km/s. A planet has twice Earth's radius and the same density. Find its escape speed.

Use the density form $v_e \propto R\sqrt{\rho}$ .
Density unchanged (factor 1); radius doubled (factor 2). So $v_e$ scales by $2 \times \sqrt{1} = 2$ .
New escape speed $= 2 \times 11.2 = 22.4$ km/s.

Answer:About 22.4 km/s.

Practice this conceptself-check · 4 quick reps

Try it yourself

Earth's escape speed is about 11.2 km/s. Another planet has HALF Earth's radius and FOUR times Earth's density. Find its escape speed.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Escape velocity is how many times the orbital speed?
2.
Same density, radius doubled. Escape speed?
3.
Escape speed in terms of g and R?
4.
Same radius, density quadrupled. Escape speed?

Halving R while quadrupling ρ leaves v_e UNCHANGED

With v_e ∝ R√ρ, a radius factor of ½ and a density factor of 4 give ½ × √4 = ½ × 2 = 1. The escape speed does not change — it stays about 11.2 km/s. The frequent slip is to combine the factors as √(½ × 2) = √1 incorrectly, or to forget that radius enters linearly while density enters as a square root.

Escape velocity is independent of the projectile's mass and launch angle

v_e = √(2GM/R) contains no reference to the escaping body's mass or the direction of launch — it is the same minimum speed for a pebble or a rocket, fired in any direction (ignoring air and obstacles).

Drill 1 more on escape velocity — v_e = √(2gm/r) and how it scales

Concept 4 of 4

What keeps a satellite up — no fuel required

Intuition

A satellite in a stable orbit is in continuous free fall: gravity bends its straight-line motion into a closed curve, and because there is no air to slow it (in space), it needs no engine, no fuel and no remote control to keep going. It simply falls around the Earth forever.

Definition

A satellite in a stable orbit needs no energy input to stay there. Gravity supplies the centripetal force, and with negligible atmospheric drag there is nothing to dissipate its energy — so it coasts indefinitely without rockets, solar power or ground control to maintain the orbit. Engines are needed only to reach orbit, change orbit, or fight residual drag — not to remain in a given orbit.

Orbit is sustained by gravity alone

F_{\text{gravity}} = \dfrac{GMm}{R^2} = \dfrac{mv_o^2}{R} \;\Rightarrow\; \text{no propulsion needed}

F_{\text{gravity}}gravitational pull = the centripetal force
v_oorbital speed

Worked example

A communications satellite has been in a stable orbit for years without firing its engines. What provides the force that keeps it on its circular path?

On a circular orbit the satellite needs a centripetal force directed toward Earth.
Earth's gravity supplies exactly this force.
With no atmosphere to slow it, no additional energy or propulsion is required to maintain the orbit.

Answer:Earth's gravity alone — the satellite needs no energy to stay in orbit.

Practice this conceptself-check · 4 quick reps

Try it yourself

Which statement about a satellite orbiting the Earth is correct: (a) it is held up by remote control, (b) retro-rockets keep it moving, (c) it needs solar panels and fuel to orbit, or (d) it requires no energy for orbiting?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
What force keeps a satellite in its circular orbit?
2.
Does a satellite need fuel to STAY in a stable orbit?
3.
Why can a satellite orbit indefinitely without engines?
4.
When does a satellite need its engines?

From the bank · past-year question

Example 4GravitationMODERATE

Which one of the following statements about a satellite orbiting around the Earth is correct?

[Q111 · Sep · 2017]

Orbiting needs no fuel — gravity does the work

A satellite is not 'held up' by rockets, remote control or solar power. It is in free fall, and gravity supplies the centripetal force. With no atmosphere to slow it, it coasts without any energy input.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Kepler's third law — T² ∝ a³
Kepler's third law
$T^2 \propto a^3 \qquad\Longleftrightarrow\qquad \dfrac{T_1}{T_2} = \left(\dfrac{a_1}{a_2}\right)^{3/2}$
Orbital velocity — v_o = √(GM/R)
Orbital velocity
$v_o = \sqrt{\dfrac{GM}{R}}$
Escape velocity — v_e = √(2GM/R) and how it scales
Escape velocity and its density scaling
$v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}; \qquad v_e \propto R\sqrt{\rho}$
What keeps a satellite up — no fuel required
Orbit is sustained by gravity alone
$F_{\text{gravity}} = \dfrac{GMm}{R^2} = \dfrac{mv_o^2}{R} \;\Rightarrow\; \text{no propulsion needed}$

Watch out for (5)

It's T² ∝ a³, not T ∝ a
→ Kepler's third law — T² ∝ a³
Orbital speed is set by the orbit, not the satellite
→ Orbital velocity — v_o = √(GM/R)
Halving R while quadrupling ρ leaves v_e UNCHANGED
→ Escape velocity — v_e = √(2GM/R) and how it scales
Escape velocity is independent of the projectile's mass and launch angle
→ Escape velocity — v_e = √(2GM/R) and how it scales
Orbiting needs no fuel — gravity does the work
→ What keeps a satellite up — no fuel required

Mastery check — 2 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1GravitationMODERATE

One year at a planet is 8 times as large as compared to one year at Earth. Which one is correct about the planet's orbit?

[Q66 · Apr · 2026]

Example 2GravitationHARD

Escape speed from the Earth is close to 11.2 km

s^{-1}

. On another planet whose radius is half of the Earth's radius and whose mass density is four times that of the Earth, the escape speed in km

s^{-1}

will be close to :

[Q84 · Apr · 2024]

Drill every past-year question on this subtopic

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