Gravitation ISC Class 11: Notes, Escape Velocity and Satellites

What Gravitation is about

Weightage: Gravitation belongs to unit 6 of the ISC Class XI Physics syllabus. CISCE weights units 4-6 together (Work, Energy and Power; Motion of System of Particles and Rigid Body; Gravitation) at 17 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Kepler's laws of planetary motion

First law (law of orbits): every planet moves in an elliptical orbit with the Sun at one focus. Second law (law of areas): the line joining the planet to the Sun sweeps out equal areas in equal intervals of time, so a planet moves fastest at perihelion (nearest point) and slowest at aphelion. This follows from conservation of angular momentum, since gravity exerts no torque about the Sun. Third law (law of periods): the square of the period is proportional to the cube of the semi-major axis, $T^2\propto a^3$. Proof for circular orbits: gravity provides the centripetal force, $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$ with $v=\frac{2\pi r}{T}$, giving $T^2=\frac{4{\pi }^2}{GM}{r}^3$.

Worked example

Mars orbits the Sun at about 1.52 times the Earth's orbital radius. Find its year in Earth years (treat both orbits as circles).

Try it first, then show the solutionSolution

1. $\frac{T_M^2}{T_E^2}=\frac{r_M^3}{r_E^3}$
2. $T_M=(1.52{)}^{3/2}\times 1year=1.87years$

Answer: About 1.87 Earth years

Newton's universal law of gravitation

Every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them: $F=\frac{G{m}_1{m}_2}{r^2}$, along the line joining them. G is the universal gravitational constant, $6.67\times 10^{-11}N{m}^{2}k{g}^{-2}$, with dimensions [M⁻¹ L³ T⁻²]; it is the same everywhere, unlike g. The force is always attractive, acts in equal and opposite pairs (third law), does not depend on the medium in between, and for uniform spheres acts as if all the mass were at the centre. It is extremely weak between everyday objects but dominates on astronomical scales.

Worked example

Find the gravitational force between two 50 kg students whose centres are 1 m apart.

Try it first, then show the solutionSolution

1. $F=\frac{6.67\times 10^{-11}\times 50\times 50}{1^2}$
2. $F=1.67\times 10^{-7}N$

Answer: About $1.7\times 10^{-7}$ N: far too small to notice

Acceleration due to gravity and its relation with G

The weight of a body of mass m at the Earth's surface is the gravitational pull of the Earth: $mg=\frac{GMm}{R^2}$, so $g=\frac{GM}{R^2}$. g is independent of the falling body's mass, so all bodies fall alike in vacuum. Writing $M=\frac{4}{3}\pi R^3\rho$ gives $g=\frac{4}{3}\pi G\rho R$. These relations let us 'weigh' the Earth: knowing g, R and G, we get $M=\frac{g{R}^2}{G}$ and the mean density $\rho =\frac{3g}{4\pi GR}$.

Worked example

Taking g = 9.8 m s⁻², R = 6.4 × 10⁶ m and G = 6.67 × 10⁻¹¹ N m² kg⁻², find the mass and mean density of the Earth.

Try it first, then show the solutionSolution

1. $M=\frac{9.8\times (6.4\times 10^6{)}^2}{6.67\times 10^{-11}}=6.0\times 10^{24}kg$
2. $\rho =\frac{3\times 9.8}{4\pi \times 6.67\times 10^{-11}\times 6.4\times 10^6}=5.5\times 10^{3}kg{m}^{-3}$

Answer: $6.0\times 10^{24}$ kg; about 5500 kg m⁻³ (more than surface rock, so the core is dense)

Variation of g with height and depth

Height h above the surface: $g_h=\frac{GM}{(R+h{)}^2}=g{\left(\frac{R}{R+h}\right)}^2$; for h much less than R, using the binomial approximation, $g_h\approx g\left(1-\frac{2h}{R}\right)$. Depth d below the surface: only the inner sphere of radius (R − d) attracts the body (the outer shell exerts no net force inside it). Its mass is $M\frac{(R-d{)}^3}{R^3}$ for uniform density, so $g_d=\frac{GM(R-d)}{R^3}=g\left(1-\frac{d}{R}\right)$. At the centre, d = R and g = 0. So g is greatest at the surface and decreases both upward (as 1/r²) and downward (linearly with r). For the same small distance, g decreases twice as fast going up as going down.

Worked example

Find g at a height of 32 km and at a depth of 32 km. (g = 9.8 m s⁻², R = 6400 km)

Try it first, then show the solutionSolution

1. Height: $g_h=9.8\left(1-\frac{2\times 32}{6400}\right)=9.8\times 0.99=9.70m{s}^{-2}$
2. Depth: $g_d=9.8\left(1-\frac{32}{6400}\right)=9.8\times 0.995=9.75m{s}^{-2}$

Answer: 9.70 m s⁻² at the height, 9.75 m s⁻² at the depth

Variation of g with latitude and rotation

The Earth spins once a day (ω = 7.27 × 10⁻⁵ rad s⁻¹). Part of the gravitational pull on a body at latitude λ is used to keep it moving in its small circle about the axis, so the effective value is $g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda$ (stated without derivation). At the poles (λ = 90°), g' = g; at the equator (λ = 0°), g' = g − ω²R, less by about 0.034 m s⁻². The Earth is also slightly flattened at the poles (polar radius smaller), which makes g at the poles larger still. If the Earth spun fast enough, with ω²R = g, bodies at the equator would be weightless.

Worked example

Find the reduction in g at the equator due to the Earth's rotation, and the length of the day for which bodies at the equator would feel weightless. (R = 6.4 × 10⁶ m, ω = 7.27 × 10⁻⁵ rad s⁻¹, g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. ${\omega }^{2}R=(7.27\times 10^{-5}{)}^2\times 6.4\times 10^6=0.034m{s}^{-2}$
2. Weightless when ${\omega }^{2}R=g$: $\omega =\sqrt{\frac{9.8}{6.4\times 10^6}}=1.24\times 10^{-3}rad{s}^{-1}$
3. T = 2π/ω = 5080 s ≈ 1.4 h

Answer: 0.034 m s⁻²; a day of about 1.4 hours

Gravitational field, potential and potential energy

Gravitational field intensity at a point is the force per unit mass on a test mass placed there: $E=\frac{GM}{r^2}$ towards the mass (N kg⁻¹, equal to m s⁻²). At the Earth's surface E = g. Gravitational potential at a point is the work done in bringing a unit mass from infinity to that point: $V=-\frac{GM}{r}$ (J kg⁻¹), negative because gravity is attractive and V = 0 at infinity. The potential difference between points at $r_A$ and $r_B$: $\Delta V=V_B-V_A=GM\left(\frac{1}{r_A}-\frac{1}{r_B}\right)$. Gravitational potential energy of mass m at distance r: $U=-\frac{GMm}{r}$. Moving from the surface to height h: $\Delta U=GMm\left(\frac{1}{R}-\frac{1}{R+h}\right)=\frac{GMmh}{R(R+h)}$, which becomes mgh when h is much less than R.

Worked example

Find the work needed to raise a 100 kg body from the Earth's surface to a height equal to the Earth's radius. (g = 9.8 m s⁻², R = 6.4 × 10⁶ m)

Try it first, then show the solutionSolution

1. $\Delta U=GMm\left(\frac{1}{R}-\frac{1}{2R}\right)=\frac{GMm}{2R}=\frac{mgR}{2}$
2. = ½ × 100 × 9.8 × 6.4 × 10⁶ = 3.14 × 10⁹ J

Answer: $3.1\times 10^9$ J (mgh would give twice this, wrongly)

Escape velocity

The escape velocity is the least speed with which a body must be projected from a planet's surface so that it never returns, reaching infinity with zero speed. Energy conservation: total energy at the surface ≥ 0, so $\frac{1}{2}m{v}_e^2-\frac{GMm}{R}=0$, giving $v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$. For the Earth, $v_e=\sqrt{2\times 9.8\times 6.4\times 10^6}$ ≈ 11.2 km s⁻¹. It does not depend on the mass of the body or the direction of projection (ignoring air). The Moon's escape velocity is only about 2.4 km s⁻¹, so gas molecules, whose fastest members approach this speed, have escaped: the Moon has no atmosphere.

Worked example

Find the escape velocity from the Moon. (g on the Moon = 1.62 m s⁻², radius 1.74 × 10⁶ m)

Try it first, then show the solutionSolution

1. $v_e=\sqrt{2gR}=\sqrt{2\times 1.62\times 1.74\times 10^6}$
2. $v_e=2.37\times 10^{3}m{s}^{-1}$

Answer: About 2.4 km s⁻¹

Orbital velocity, period and energy of a satellite

For a satellite of mass m in a circular orbit of radius r = R + h, gravity supplies the centripetal force: $\frac{m{v}_o^2}{r}=\frac{GMm}{r^2}$, so $v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{g{R}^2}{R+h}}$, independent of m. Near the surface (h ≪ R), $v_o=\sqrt{gR}$ ≈ 7.9 km s⁻¹, so $v_e=\sqrt{2}{v}_o$. Period: $T=\frac{2\pi r}{v_o}=2\pi \sqrt{\frac{r^3}{GM}}$. Energies: KE = $\frac{GMm}{2r}$, PE = $-\frac{GMm}{r}$, total E = $-\frac{GMm}{2r}$; the negative total shows the satellite is bound, and $\frac{GMm}{2r}$ is its binding energy. Weightlessness: the satellite and everything in it fall towards the Earth with the same acceleration, so there is no normal reaction between an astronaut and the floor; the apparent weight is zero, although gravity still acts.

Worked example

A satellite orbits 600 km above the Earth's surface. Find its orbital speed and period. (g = 9.8 m s⁻², R = 6400 km)

Try it first, then show the solutionSolution

1. r = 7.0 × 10⁶ m; $v_o=\sqrt{\frac{9.8\times (6.4\times 10^6{)}^2}{7.0\times 10^6}}=7.57\times 10^{3}m{s}^{-1}$
2. $T=\frac{2\pi \times 7.0\times 10^6}{7.57\times 10^3}=5.81\times 10^{3}s$ ≈ 97 min

Answer: 7.57 km s⁻¹; about 97 minutes

Geostationary and polar satellites

A geostationary satellite appears fixed above one point on the equator. It must (1) orbit in the equatorial plane, (2) move in the same direction as the Earth spins (west to east) and (3) have a period of 24 hours. From $T=2\pi \sqrt{\frac{r^3}{GM}}$, the radius of this parking orbit is $r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{1/3}$ ≈ 4.2 × 10⁷ m, a height of about 36 000 km. Uses: communication, television broadcasting and weather watching over a fixed region; three such satellites spaced 120° apart cover almost the whole Earth. Polar satellites orbit at low heights (about 500-800 km) in planes passing over the poles; as the Earth turns beneath them they scan the entire surface strip by strip, which suits remote sensing, mapping and weather forecasting.

Worked example

Find the radius of the geostationary orbit and its height above the surface. (g = 9.8 m s⁻², R = 6.4 × 10⁶ m, T = 86 400 s)

Try it first, then show the solutionSolution

1. GM = gR² = 9.8 × (6.4 × 10⁶)² = 4.01 × 10¹⁴ m³ s⁻²
2. $r={\left(\frac{4.01\times 10^{14}\times 86400^2}{4{\pi }^2}\right)}^{1/3}=4.23\times 10^{7}m$
3. h = r − R = 3.59 × 10⁷ m

Answer: About $4.2\times 10^7$ m from the centre, roughly 36 000 km above the surface

Physics in real life: Gravitation

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Acceleration Due to Gravity and Its Variation
• Escape Velocity and Satellite Motion

Gravitation formula sheet

Universal law of gravitation

$F=\frac{G{m}_1{m}_2}{r^2},G=6.67\times 10^{-11}N{m}^{2}k{g}^{-2}$

[G] = [M⁻¹ L³ T⁻²].

g and G

$g=\frac{GM}{R^2}=\frac{4}{3}\pi G\rho R$

Gives M = gR²/G.

g at height

$g_h=g{\left(\frac{R}{R+h}\right)}^2\approx g\left(1-\frac{2h}{R}\right)$

Approximation for h ≪ R.

g at depth

$g_d=g\left(1-\frac{d}{R}\right)$

Zero at the centre.

Latitude and rotation

$g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda$

No derivation; least at the equator.

Field and potential

$E=\frac{GM}{r^2},V=-\frac{GM}{r},\Delta V=GM\left(\frac{1}{r_A}-\frac{1}{r_B}\right)$

V = 0 at infinity.

Potential energy

$U=-\frac{GMm}{r};\Delta U=mgh(h\ll R)$

Use −GMm/r for large heights.

Escape velocity

$v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\approx 11.2km{s}^{-1}$

Independent of the body's mass.

Orbital velocity

$v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{g{R}^2}{R+h}};v_e=\sqrt{2}{v}_o(\text{near surface})$

≈ 7.9 km s⁻¹ near the surface.

Period and Kepler's third law

$T=2\pi \sqrt{\frac{r^3}{GM}},T^2\propto r^3$

Geostationary: T = 24 h, h ≈ 36 000 km.

Satellite energy

$K=\frac{GMm}{2r},E=-\frac{GMm}{2r}$

Binding energy = GMm/2r.

Gravitation: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Gravitation

The question patterns to expect, and the method that works for each.

Tricks to remember Gravitation

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Kepler in three words

  Ellipse, Equal areas, T² ∝ r³: 'E-E-T'.

  Use it when: Stating Kepler's laws in order.

• Up twice, down once

  For small distances, g falls by 2h/R going up but only d/R going down.

  Use it when: Comparing g at height and depth.

• Root two between them

  Escape speed = √2 × orbital speed near the surface: 7.9 × 1.414 ≈ 11.2 km s⁻¹.

  Use it when: Linking v_e and v_o.

• Satellite energy halves

  KE = +GMm/2r, PE = −GMm/r, total = −GMm/2r: total is minus the KE.

  Use it when: Satellite energy questions.

• GEO: 24, equator, eastward

  A geostationary satellite needs 24 h, the equatorial plane and west-to-east motion.

  Use it when: Conditions for a geostationary satellite.

Gravitation: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Gravitation

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using mgh for large heights

  mgh assumes g is constant; for heights comparable with R it overestimates the energy.

  Wrong working

  Work to raise 100 kg to height R: mgR = 6.3 × 10⁹ J.

  Right working

  ΔU = GMm(1/R − 1/2R) = mgR/2 = 3.1 × 10⁹ J.

  ✓Do this: Use −GMm/r at both ends unless h ≪ R.

  Usually costs: 1-2 marks.

• ✗Mistake: Using h instead of R + h in orbital formulae

  Students put the height above the surface into v = √(GM/r).

  Wrong working

  Satellite 600 km up: r = 6.0 × 10⁵ m.

  Right working

  r = R + h = 7.0 × 10⁶ m.

  ✓Do this: Always measure r from the Earth's centre.

  Usually costs: 1-2 marks.

• ✗Mistake: Using the small-height approximation when h is not small

  g(1 − 2h/R) gives nonsense for h = R (it would give −g).

  Wrong working

  At h = R: g_h = g(1 − 2) = −g.

  Right working

  g_h = g(R/(R + h))² = g/4.

  ✓Do this: Use the approximation only when h is a few per cent of R.

  Usually costs: 1 mark.

• ✗Mistake: Saying there is no gravity in an orbiting satellite

  Students explain weightlessness as 'g = 0 in space'.

  Wrong working

  Astronauts are weightless because there is no gravity at that height.

  Right working

  Gravity is still about 90% of its surface value 400 km up; astronauts feel weightless because they and the satellite fall freely together.

  ✓Do this: Link weightlessness to free fall and zero normal reaction.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting the conditions for a geostationary orbit

  Students give only '24 hour period'.

  Wrong working

  A geostationary satellite has a period of one day.

  Right working

  Period 24 h, orbit in the equatorial plane, moving west to east (same sense as the Earth's spin).

  ✓Do this: List all three conditions.

  Usually costs: 1 mark.

Practice questions on Gravitation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The SI unit of G is:
   (a) N m kg⁻²
   (b) N m² kg⁻²
   (c) N kg² m⁻²
   (d) N m² kg

   Hint

   G = Fr²/m₁m₂.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   The value of g at the centre of the Earth is:
   (a) 9.8 m s⁻²
   (b) infinite
   (c) zero
   (d) 4.9 m s⁻²

   Hint

   Put d = R.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   State Kepler's law of areas.

   Hint

   Equal areas.

   Answer

   The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

4. Q4 · 1 mark · Very short answerExam level

   State the conditions for a satellite to be geostationary.

   Hint

   Period, plane, direction.

   Answer

   Period 24 h, orbit in the equatorial plane, moving west to east.

5. Q5 · 2 marks · Short answerExam level

   Show that near the Earth's surface the escape velocity is √2 times the orbital velocity.

   Hint

   Compare √(2gR) and √(gR).

   Answer

   $v_e=\sqrt{2GM/R}$ and $v_o=\sqrt{GM/R}$ (r ≈ R), so $v_e/v_o=\sqrt{2}$.

6. Q6 · 2 marks · Short answerHarder exam level

   Find the ratio of g at a height R above the surface to g at a depth R/2 below it.

   Hint

   g/4 and g/2.

   Answer

   g/4 : g/2 = 1 : 2.

7. Q7 · 2 marks · Short answerHarder exam level

   If the Earth's radius decreased by 1% with no change in mass, how would g change?

   Hint

   g ∝ R⁻².

   Answer

   Δg/g = −2ΔR/R = +2%: g increases by about 2%.

8. Q8 · 2 marks · Short answerHarder exam level

   Find the gravitational potential energy of a 10 kg body at a height R above the Earth's surface. (g = 9.8 m s⁻², R = 6.4 × 10⁶ m)

   Hint

   U = −GMm/2R = −mgR/2.

   Answer

   U = −10 × 9.8 × 6.4 × 10⁶/2 = −3.1 × 10⁸ J.

9. Q9 · 2 marks · Short answerHarder exam level

   The periods of two planets are in the ratio 1 : 8. Find the ratio of their orbital radii.

   Hint

   r ∝ T^(2/3).

   Answer

   r₁ : r₂ = (1/8)^(2/3) = 1 : 4.

10. Q10 · 3 marks · Short answerHarder exam level

    Two 100 kg spheres have their centres 2 m apart. Find the gravitational field and the potential at the midpoint between them. (G = 6.67 × 10⁻¹¹ N m² kg⁻²)

    Hint

    Field is a vector; potential is a scalar.

    Answer

    Fields are equal and opposite, so E = 0. V = 2 × (−G × 100/1) = −1.33 × 10⁻⁸ J kg⁻¹.

11. Q11 · 3 marks · Short answerHarder exam level

    Find the orbital speed and period of a satellite at a height equal to half the Earth's radius. (g = 9.8 m s⁻², R = 6.4 × 10⁶ m)

    Hint

    r = 1.5R.

    Answer

    v = √(gR²/1.5R) = √(gR/1.5) = 6.47 km s⁻¹; T = 2π(9.6 × 10⁶)/6466 = 9.33 × 10³ s ≈ 2.6 h.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) Define gravitational potential at a point and write its expression for a point at distance r from a mass M. [1]
    (b) Show that the change in potential energy in raising a body of mass m through a small height h is mgh. [2]
    (c) Find the height of a geostationary satellite above the Earth's surface. (g = 9.8 m s⁻², R = 6.4 × 10⁶ m) [2]

    Hint

    ΔU = GMmh/R(R + h).

    Answer

    (a) The work done in bringing unit mass from infinity to the point: V = −GM/r. (b) ΔU = GMm(1/R − 1/(R + h)) = GMmh/(R(R + h)) ≈ GMmh/R² = mgh for h ≪ R. (c) r = (gR²T²/4π²)^(1/3) = 4.23 × 10⁷ m; h ≈ 3.6 × 10⁷ m (about 36 000 km).

13. Q13 · 3 marks · Short answerHOTS / competency

    A 200 kg satellite circles the Earth just above the surface (r ≈ R). By what percentage must its speed be increased for it to escape, and how much extra kinetic energy must its engines supply? (g = 9.8 m s⁻², R = 6.4 × 10⁶ m)

    Hint

    Compare √(gR) with √(2gR).

    Answer

    $v_o=\sqrt{gR}$ and $v_e=\sqrt{2gR}=\sqrt{2}{v}_o$, so the speed must rise by about 41.4%. Extra KE = ½m(2gR − gR) = ½mgR = ½ × 200 × 9.8 × 6.4 × 10⁶ = $6.27\times 10^9$ J (equal to its KE in orbit: the total energy −½mgR must be raised to zero).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State Kepler's third law.Show answer

   T² ∝ r³ (the cube of the semi-major axis).

2. 2.Dimensions of G?Show answer

   [M⁻¹ L³ T⁻²].

3. 3.g at depth d?Show answer

   g(1 − d/R).

4. 4.Where on Earth is g greatest?Show answer

   At the poles.

5. 5.Expression for gravitational potential?Show answer

   V = −GM/r.

6. 6.Escape velocity from the Earth?Show answer

   About 11.2 km s⁻¹ (√(2gR)).

7. 7.Height of a geostationary satellite?Show answer

   About 36 000 km.

8. 8.Is the Cavendish experiment in the ISC syllabus?Show answer

   No.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state Kepler's laws and prove the third for circular orbits?
• ☐Can you find the mass of the Earth from g?
• ☐Can you derive g at height h and depth d?
• ☐Can you explain why g is least at the equator?
• ☐Can you define gravitational potential and write V and U?
• ☐Can you derive the escape velocity?
• ☐Can you find the speed, period and energy of a satellite?
• ☐Can you find the height of a geostationary orbit and give its conditions?

Gravitation: frequently asked questions