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  3. Satellite Motion – Orbital Velocity, Time Period and Energy

Gravitation · JEE & NEET Physics

Satellite Motion – Orbital Velocity, Time Period and Energy: notes and previous year questions

How fast satellites move, how long they take to go round, their energy (1 : 2 : 1), launch energy, geostationary and other orbits, and weightlessness.

Satellite Motion – Orbital Velocity, Time Period and Energy in short

  • A satellite is always falling and always missing: gravity provides the centripetal force.
  • Orbital speed v = √(GM/r) does not depend on the satellite's mass; higher is slower.
  • Just above the ground: v ≈ 7.9 km/s and T ≈ 84 min, the shortest possible period.
  • T = 2π√(r³/GM), so T² ∝ r³.

1Always falling

A space station is not held up: it is falling. In one second it moves 7.7 km sideways and falls about 4.4 m, but over 7.7 km the curved ground also drops away by about 4.4 m, so it never gets closer.

A satellite is any body that goes round a planet, held by gravity: the Moon, or weather, TV and navigation satellites. Gravity provides exactly the centripetal force, so no engine is needed to stay in orbit.

2Orbital velocity

GMmr2=mv2r ⇒ vo=GMr=gR2R+h\frac{GMm}{r^2} = \frac{mv^2}{r}\ \Rightarrow\ v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{gR^2}{R+h}}
  • Just above the ground: vo=gR≈7.9v_o = \sqrt{gR} \approx 7.9 km/s (8 km/s with g=10g = 10).
  • It does not depend on the satellite's mass.
  • Higher is slower: v∝1/rv \propto 1/\sqrt r. At h=Rh = R: 7.9/2≈5.67.9/\sqrt2 \approx 5.6 km/s; at r=3Rr = 3R: about 4.6 km/s.
  • Orbits at rr and 4r4r: speeds 2:12 : 1. Half the speed needs 4 times the radius.
  • From the same place, the escape speed is 2 vo\sqrt2\,v_o.

3Time period

T=2πrv=2πr3GM=2πr3gR2T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{r^3}{gR^2}}T² ∝ r³: Kepler's third law.

Near the ground: T0=2πR/g=2π×800≈5027T_0 = 2\pi\sqrt{R/g} = 2\pi \times 800 \approx 5027 s ≈84\approx 84 min, the shortest possible orbit round the Earth. At h=Rh = R: T=22×84≈237T = 2\sqrt2 \times 84 \approx 237 min, nearly 4 hours.

SatelliteHeightPeriod
Just above the ground084 min
Space station≈ 400 km≈ 92 min
GPS≈ 20,200 km12 h
Geostationary≈ 35,800 km24 h
The Moon≈ 384,000 km27.3 days

4Energy of a satellite

KE=12mv2=GMm2rKE = \tfrac12 mv^2 = \frac{GMm}{2r}
PE=−GMmrPE = -\frac{GMm}{r}
E=KE+PE=−GMm2rE = KE + PE = -\frac{GMm}{2r}Negative: the satellite is bound.

So KE:∣PE∣:∣E∣=1:2:1KE : |PE| : |E| = 1 : 2 : 1, with E=−KE=PE/2E = -KE = PE/2. If KE=100KE = 100 J, then PE=−200PE = -200 J and E=−100E = -100 J. For 100 kg at h=Rh = R (g=10g = 10): KE=1.6×109KE = 1.6\times10^9 J, PE=−3.2×109PE = -3.2\times10^9 J, E=−1.6×109E = -1.6\times10^9 J.

5Energy to launch

ΔE=GMm(1R−12r)=mgR(1−R2(R+h))\Delta E = GMm\Big(\frac1R - \frac1{2r}\Big) = mgR\Big(1 - \frac{R}{2(R+h)}\Big)From rest on the ground (E = −GMm/R) to a circular orbit at r = R + h.

For a low orbit, ΔE≈12mgR\Delta E \approx \tfrac12 mgR: half the energy needed to escape. For h=Rh = R: ΔE=34GMm/R\Delta E = \tfrac34 GMm/R, three times the final KE of 14GMm/R\tfrac14 GMm/R (the rest goes into PE).

6Special orbits

A geostationary satellite stays above one spot, so a dish never has to move. It needs all three:

  1. a period of one day: one turn of the Earth relative to the stars, 23 h 56 min;
  2. an orbit in the plane of the equator;
  3. west-to-east motion, the same way the Earth turns.

From T∝r3/2T \propto r^{3/2}: r=(1436/84.6)2/3R≈6.6R≈42,200r = (1436/84.6)^{2/3}R \approx 6.6R \approx 42{,}200 km, a height of about 35,800 km, at about 3.1 km/s. A geosynchronous orbit has the one-day period but is tilted, so it drifts north and south.

OrbitHeightPeriodUsed for
Low Earth orbit200–2000 km90–120 minspace station, cameras; air drag slowly pulls it down
Polar500–800 km≈ 100 minmaps, weather: sees the whole Earth as it turns
GPS (medium)≈ 20,200 km12 hnavigation
Geostationary≈ 35,800 km24 hTV, weather, phones

7Weightlessness

At 400 km, g=9.8×(6400/6800)2≈8.7g = 9.8 \times (6400/6800)^2 \approx 8.7 m/s², nearly 90% of its value on the ground. Astronauts float because the station, the astronaut and every loose object fall together with the same acceleration: the floor never pushes, so the apparent weight is zero.

Summary

Key ideas

  • A satellite is always falling and always missing: gravity provides the centripetal force.
  • Orbital speed v = √(GM/r) does not depend on the satellite's mass; higher is slower.
  • Just above the ground: v ≈ 7.9 km/s and T ≈ 84 min, the shortest possible period.
  • T = 2π√(r³/GM), so T² ∝ r³.
  • KE = GMm/2r, PE = −GMm/r, E = −GMm/2r: the ratio 1 : 2 : 1.
  • A higher orbit has less KE but more total energy.
  • Launching into a low orbit needs half the escape energy.
  • Geostationary: one sidereal day, over the equator, west to east, about 35,800 km up.
  • Weightless means falling together, not zero gravity.

Every equation

Orbital speed
vo=GM/rv_o = \sqrt{GM/r}
With g and R
vo=gR2/(R+h)v_o = \sqrt{gR^2/(R+h)}
Just above the ground
vo=gRv_o = \sqrt{gR}
Period
T=2πr3/GMT = 2\pi\sqrt{r^3/GM}
Shortest period
T0=2πR/gT_0 = 2\pi\sqrt{R/g}
Kinetic energy
KE=GMm/2rKE = GMm/2r
Potential energy
PE=−GMm/rPE = -GMm/r
Total energy
E=−GMm/2rE = -GMm/2r
Energy ratio
KE:∣PE∣:∣E∣=1:2:1KE : |PE| : |E| = 1 : 2 : 1
Elliptical orbit
E=−GMm/2aE = -GMm/2a
Launch energy
ΔE=GMm(1/R−1/2r)\Delta E = GMm(1/R - 1/2r)
Geostationary radius
r≈6.6R≈42,200 kmr \approx 6.6R \approx 42{,}200\ \text{km}
g at a height
gh=g (R/(R+h))2g_h = g\,(R/(R+h))^2

Previous year questions with solutions

Real JEE and NEET questions on satellite motion – orbital velocity, time period and energy. Try each one before you open the solution.

Q1NEET 2026One correct option

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius RR is proportional to:

  1. AR3R^{3}
  2. BR1/2R^{1/2}
  3. CR3/2R^{3/2}
  4. DR2R^{2}
Show answer and solution

Answer: Option C

From the last question, T=2πr3GMT = 2\pi\sqrt{\dfrac{r^{3}}{GM}}, with MM now the Sun's mass. Everything but the radius is the same for every planet of the system, so T∝R3=R3/2T \propto \sqrt{R^{3}} = R^{3/2}; squared, T2∝R3T^{2} \propto R^{3}.

The trap is A, R3R^{3}, which is how T2T^{2} goes, not TT. B, R1/2R^{1/2}, is how the orbital speed goes, turned upside down: vo∝R−1/2v_{o} \propto R^{-1/2}. D, R2R^{2}, comes from no correct working. Check: a planet four times as far out takes 43/2=84^{3/2} = 8 times as long.

Q2JEE Main 2026One correct option

A planet (P1P_{1}) is moving around the star of mass 2M2M in the orbit of radius RR. Another planet (P2P_{2}) is moving around another star of mass 4M4M in a orbit of radius 2R2R. Ratio of time periods of revolution of P2P_{2} and P1P_{1} is ________.

  1. A12\frac{1}{2}
  2. B22
  3. C44
  4. D14\frac{1}{4}
Show answer and solution

Answer: Option B

T∝r3MT \propto \sqrt{\dfrac{r^{3}}{M}}, so T2T1=(2R)34M×2MR3=8×24=4=2\dfrac{T_{2}}{T_{1}} = \sqrt{\dfrac{(2R)^{3}}{4M} \times \dfrac{2M}{R^{3}}} = \sqrt{\dfrac{8 \times 2}{4}} = \sqrt{4} = 2.

The trap is A, 12\dfrac{1}{2}, which is T1T2\dfrac{T_{1}}{T_{2}}, the ratio upside down. C, 44, forgets the square root: it is the ratio of the squares of the periods. D, 14\dfrac{1}{4}, makes both slips. Check in two steps: the orbit twice as wide alone multiplies the period by 222\sqrt{2}, and the star twice as heavy divides it by 2\sqrt{2}, leaving 22.

Q3JEE Advanced 2021Numerical answer

The distance between two stars of masses 3M_{S} and 6M_{S} is 9R. Here R is the mean distance between the centers of the Earth and the Sun, and M_{S} is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period nT, where T is the period of Earth's revolution around the Sun. The value of n is __________.

Show answer and solution

Answer: 9

For the Earth round the Sun, T2=4π2R3GMST^{2} = \dfrac{4\pi^{2}R^{3}}{GM_{S}} (the Earth's mass is tiny beside the Sun's). For the binary, the separation is 9R9R and the total mass 3MS+6MS=9MS3M_{S} + 6M_{S} = 9M_{S}:

(nT)2=4π2(9R)3G⋅9MS=7299⋅4π2R3GMS=81T2(nT)^{2} = \dfrac{4\pi^{2}(9R)^{3}}{G \cdot 9M_{S}} = \dfrac{729}{9} \cdot \dfrac{4\pi^{2}R^{3}}{GM_{S}} = 81T^{2}

so n=9n = 9.

The trap is using one star's mass instead of the sum: with 6MS6M_{S} alone, n2=7296n^{2} = \dfrac{729}{6} and n≈11n \approx 11; with 3MS3M_{S}, n≈15.6n \approx 15.6. Another is using the radius of one star's circle instead of the separation: the law has the full distance between the stars.

Practice questions, easy to hard

Three questions from the satellite motion – orbital velocity, time period and energy practice ladder: one easy, one medium, one hard.

Q4One or more correct options

A satellite's angular momentum about the planet's centre is L=mvorL = mv_{o}r, since its velocity is at right angles to the radius. With vo=GMrv_{o} = \sqrt{\dfrac{GM}{r}},

L=mGMr r=mGMrL = m\sqrt{\dfrac{GM}{r}}\, r = m\sqrt{GMr}

So a satellite in a wider orbit carries more angular momentum, though it moves more slowly.

Using its engines, a satellite is moved from a circular orbit of radius rr to one of radius 4r4r round the same planet. Which statements are correct?

  1. AIts speed is halved
  2. BIts period becomes 88 times as long
  3. CIts angular momentum doubles
  4. DIts angular momentum is unchanged, because gravity pulls towards the centre and exerts no torque
Show answer and solution

Answer: Options A, B, C

A: vo∝1rv_{o} \propto \dfrac{1}{\sqrt{r}}, so four times the radius halves the speed. B: T∝r3/2T \propto r^{3/2}, and 43/2=84^{3/2} = 8. C: L=mGMr∝rL = m\sqrt{GMr} \propto \sqrt{r}, and 4=2\sqrt{4} = 2. Check with L=mvrL = mvr: half the speed at four times the radius is twice the product.

D is the trap. Gravity alone does keep LL fixed while the satellite coasts, but the move to the new orbit was made by the engines, whose push is not towards the centre. Their torque is what changed LL.

Q5Numerical answer

The result v∞=v2−ve2v_{\infty} = \sqrt{v^{2} - v_{e}^{2}} needs no GG, no MM and no RR, only the escape velocity of the place the body leaves from.

A probe is launched from the Earth at 14 km/s14\ \mathrm{km/s}. The Earth's escape velocity is 11.2 km/s11.2\ \mathrm{km/s}. Ignoring air and every other body, what is its speed very far from the Earth, in km/s\mathrm{km/s}? Give one decimal place.

Show answer and solution

Answer: 8.4 km/s

v∞2=142−11.22=196−125.44=70.56v_{\infty}^{2} = 14^{2} - 11.2^{2} = 196 - 125.44 = 70.56, so v∞=8.4 km/sv_{\infty} = 8.4\ \mathrm{km/s}.

The trap is 2.8 km/s2.8\ \mathrm{km/s}, from 14−11.214 - 11.2. It looks natural, but it is the squares that subtract, and the probe keeps far more than the bare margin of its launch speed over vev_{e}.

Q6Numerical answer

Each star is held on its circle by the other's pull. For star 1, Gm1m2d2=m1ω2r1\dfrac{Gm_{1}m_{2}}{d^{2}} = m_{1}\omega^{2}r_{1}. Put in r1=m2dm1+m2r_{1} = \dfrac{m_{2}d}{m_{1} + m_{2}} and m2m_{2} cancels, giving ω2=G(m1+m2)d3\omega^{2} = \dfrac{G(m_{1} + m_{2})}{d^{3}}, so

T=2πd3G(m1+m2)T = 2\pi\sqrt{\dfrac{d^{3}}{G(m_{1} + m_{2})}}

This is the one-planet law with the separation dd in place of rr and the total mass in place of MM.

Two equal stars, each of mass MM, orbit their centre of mass a distance dd apart. A light planet circles a single star of mass MM at radius dd. Find TbinaryTplanet\dfrac{T_{\text{binary}}}{T_{\text{planet}}} to two decimal places.

Show answer and solution

Answer: 0.71

Both have the same d3d^{3}; the binary's total mass is 2M2M, the planet's star MM. So TbinaryTplanet=M2M=12≈0.71\dfrac{T_{\text{binary}}}{T_{\text{planet}}} = \sqrt{\dfrac{M}{2M}} = \dfrac{1}{\sqrt{2}} \approx 0.71.

The trap is 11: the second star is not just a passenger, and its pull adds to the first's. Another is 2≈1.41\sqrt{2} \approx 1.41, the ratio upside down: more mass pulling means a shorter period, not a longer one. And 0.50.5 forgets the square root.