MHT-CET Physics · Gravitation

Satellites, Orbits and Kepler's Laws

A satellite in a circular orbit of radius r has gravity as its centripetal force, so its speed is √(GM/r), its period grows as r^(3/2) (Kepler's third law, T² ∝ r³), and its total energy −GMm/2r is half its potential energy.

All Gravitation notes MHT-CET Physics strategy

Why this matters

21 PYQs, 6 of them HARD. Seventeen scale the orbital speed or period with the radius — two satellites compared, a geostationary orbit, a satellite skimming a planet of given density, a force law other than inverse square, a comet's aphelion. Four are a satellite's energy: total energy, the energy to launch it, and ratios. Two cards.

Concept 1 of 2: Orbital Speed, Period and Kepler's Third Law

Setting GMm/r² equal to mv²/r gives v = √(GM/r): four times the radius, half the speed. The period 2πr/v = 2π√(r³/GM) grows as r^(3/2), so T² ∝ r³ — a period 8 times longer means an orbit 4 times larger, and a quarter the radius means an eighth of the period. Just above a planet of density ρ, T = √(3π/Gρ), independent of the planet's size. A geostationary orbit has the earth's own ω: r³ = gR²/ω². The angular momentum mvr = m√(GMr) grows as √r. If the force followed some other power, F ∝ r⁻ᵏ, the same balance gives T² ∝ r^(k+1). Two equal masses circling their midpoint at radius r are 2r apart, so v = √(Gm/4r).

Definition

$v = \sqrt{\dfrac{GM}{r}}$ , $T = 2\pi\sqrt{\dfrac{r^3}{GM}}$ , $T^2 \propto r^3$ .
At height R: $T = 4\pi\sqrt{\dfrac{2R}{g}}$ . Close to a planet: $T = \sqrt{\dfrac{3\pi}{G\rho}}$ , $T^2\rho = \dfrac{3\pi}{G}$ .
Geostationary: $r = \left(\dfrac{gR^2}{\omega^2}\right)^{1/3}$ . Angular momentum $L = m\sqrt{GMr} \propto r^{1/2}$ .
Force ∝ $r^{-k}$ : $T^2 \propto r^{k+1}$ (k = 7/2 ⇒ $r^{9/2}$ ).
Stable circular orbit needs the horizontal speed to equal the critical speed.

Circular orbit

v = \sqrt{\frac{GM}{r}}, \qquad T = 2\pi\sqrt{\frac{r^3}{GM}}

Worked example

Two satellites orbit at radii R and 9R. Ratio of their speeds and of their periods?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q26Hard

Example 1 · Gravitation · Satellites, Orbital Motion, and Kepler's Laws

Periodic time of a satellite revolving above the earth's surface at a height equal to radius of the earth

R

is (

g

= acceleration due to gravity)

Using the height instead of the orbit radius

A satellite at height R above the surface orbits at radius 2R. The period is 2π√((2R)³/gR²), not 2π√(R/g).

Taking the separation as the orbit radius

Two equal masses circling their midpoint are 2r apart. The force uses (2r)², the circular motion uses r.

Drill 16 more on orbital speed, period and kepler's third law

Concept 2 of 2: A Satellite's Energy

In orbit, KE = GMm/2r, PE = −GMm/r and total E = −GMm/2r: the total is half the potential energy and minus the kinetic energy. Launching from the surface to an orbit of radius r needs the difference between the orbit's total energy and the surface value −GMm/R. The energy scales as m/r, so a satellite three times heavier at a quarter the radius has twelve times the energy.

Definition

$KE = \dfrac{GMm}{2r}$ , $PE = -\dfrac{GMm}{r}$ , $E = -\dfrac{GMm}{2r} = \dfrac{PE}{2} = -KE$ .
Launch from the surface to altitude 2R (r = 3R): $\Delta E = \dfrac{GMm}{R} - \dfrac{GMm}{6R} = \dfrac{5GMm}{6R}$ .
$|E| \propto \dfrac{m}{r}$ : masses 3 : 1 at r and 4r ⇒ 12 : 1.

Orbital energy

E = -\frac{GMm}{2r} = \frac{U}{2} = -K

Worked example

Minimum energy to put a satellite of mass m from the surface into orbit at altitude R?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q5Moderate

Example 2 · Gravitation · Satellites, Orbital Motion, and Kepler's Laws

The minimum energy required to launch a satellite of mass

m

from the surface of a planet of mass

M

and radius

R

in a circular orbit at an altitude of

2R

Launching with only the orbit's kinetic energy

From the surface the satellite must also be lifted. The launch energy is the total orbital energy minus the surface energy −GMm/R, not GMm/2r alone.

Drill 3 more on a satellite's energy

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 (2)

Orbital Speed, Period and Kepler's Third Law
Circular orbit
$v = \sqrt{\frac{GM}{r}}, \qquad T = 2\pi\sqrt{\frac{r^3}{GM}}$
A Satellite's Energy
Orbital energy
$E = -\frac{GMm}{2r} = \frac{U}{2} = -K$

Watch out for (3)

Using the height instead of the orbit radius
→ Orbital Speed, Period and Kepler's Third Law
Taking the separation as the orbit radius
→ Orbital Speed, Period and Kepler's Third Law
Launching with only the orbit's kinetic energy
→ A Satellite's Energy

Test yourself on Gravitation

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Gravitation chapter test Drill all 21 questions