Class 11 Physics · Original EZT practice
Gravitation (Class 11 Chapter 7): summary and practice
Gravitation takes Newton's inverse-square law from falling apples to orbiting satellites: Kepler's laws, the gravitational constant, how g changes below and above the Earth's surface, escape speed, and the energy of an orbiting satellite. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Kepler's laws
Planets sweep out ellipses with the Sun at one focus, cover equal areas in equal times, and obey T^2 proportional to a^3 - the square of the period grows with the cube of the orbit's semi-major axis. Newton later showed all three follow from one inverse-square force.
2. The universal law and G
Every mass attracts every other: F = G m1 m2 / r^2 along the line joining them, with G = 6.67 x 10^-11 N m^2/kg^2 measured by Cavendish-style experiments. The force is tiny between lab objects and enormous between planets.
3. g below and above the surface
At the surface g = G M / R^2, about 9.81 m/s^2 for Earth. Above the surface g falls as 1/(R + h)^2 - the simple g(1 - 2h/R) shortcut holds only when h is much smaller than R. Below the surface only the mass inside your radius pulls, so g decreases linearly toward zero at the centre.
4. Potential energy and escape speed
Gravitational potential energy is U = -G M m / r, negative because zero is set at infinity. Escape speed is the launch speed that just reaches infinity: v = sqrt(2 G M / R) = sqrt(2 g R), about 11.2 km/s for Earth, and it does not depend on the projectile's mass.
5. Earth satellites and orbital energy
A satellite in a circular orbit of radius r moves at v = sqrt(G M / r) with period T = 2 pi sqrt(r^3 / G M). Its total energy is -G M m / 2r: half the potential energy, with the kinetic energy positive and exactly cancelling half. Geostationary satellites park above the equator with a 24-hour period.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Apply the universal law
Choose the method: F = G m1 m2 / r^2.
EZT practice question
Two 100 kg spheres sit with centres 1 m apart. Find the gravitational force between them (G = 6.67 x 10^-11 N m^2/kg^2).
Show the worked solution
- F = G m1 m2 / r^2.
- = 6.67 x 10^-11 x 100 x 100 / 1.
Answer: F = 6.67 x 10^-7 N - less than a micronewton, which is why lab-scale gravity needs delicate experiments.
Check: Gravitation is always attractive and acts along the line joining the centres.
Watch for this: r is the CENTRE-TO-CENTRE distance, not the gap between the surfaces.
Now try it yourself
10 kg and 20 kg masses with centres 2 m apart. Force?
Check your answer
3.35 x 10^-9 N (approximately).
6.67 x 10^-11 x 200 / 4 = 3.34 x 10^-9 N.
Read g at the surface
Choose the method: g = G M / R^2.
EZT practice question
Using G = 6.67 x 10^-11, M = 6 x 10^24 kg and R = 6.4 x 10^6 m, estimate g at the Earth's surface.
Show the worked solution
- g = G M / R^2.
- = 6.67 x 10^-11 x 6 x 10^24 / (6.4 x 10^6)^2.
- = 4.00 x 10^14 / 4.10 x 10^13.
Answer: g = 9.81 m/s^2 (approximately).
Check: Every planet has its own g because M and R differ - the Moon's is about a sixth of Earth's.
Watch for this: Square the radius BEFORE dividing - the most common slip in this formula.
Now try it yourself
A planet with the same mass but twice Earth's radius. Surface g?
Check your answer
About 2.45 m/s^2.
g scales as 1/R^2, so doubling R quarters g: 9.81/4 = 2.45 m/s^2.
Compute g at a height
Choose the method: g(h) = G M / (R + h)^2; the shortcut g(1 - 2h/R) works only for h much smaller than R.
EZT practice question
Find g at a height of 400 km above the surface (R = 6400 km, g = 9.81 m/s^2).
Show the worked solution
- g(h) = g R^2 / (R + h)^2.
- = 9.81 x (6400/6800)^2.
- = 9.81 x 0.8858.
Answer: g = 8.69 m/s^2 (approximately).
Check: The small-height approximation would give 9.81 x (1 - 800/6400) = 8.58 m/s^2 - close but off, because 400 km is not negligible against 6400 km.
Watch for this: 400 km is 6 percent of R - use the exact (R + h)^2 formula, not the linear shortcut.
Now try it yourself
g at h = R (6400 km up)?
Check your answer
About 2.45 m/s^2.
(R/2R)^2 = 1/4, so g falls to a quarter: 9.81/4.
Escape speed
Choose the method: v_escape = sqrt(2 G M / R) = sqrt(2 g R).
EZT practice question
Compute Earth's escape speed with g = 9.81 m/s^2 and R = 6.4 x 10^6 m.
Show the worked solution
- v = sqrt(2 g R).
- = sqrt(2 x 9.81 x 6.4 x 10^6).
- = sqrt(1.256 x 10^8).
Answer: v = 1.12 x 10^4 m/s, about 11.2 km/s.
Check: Escape speed is independent of the projectile's mass - a cricket ball and a rocket both need 11.2 km/s from the surface.
Watch for this: Do not halve the formula - sqrt(g R) is the orbital speed at the surface, sqrt(2 g R) is escape.
Now try it yourself
The Moon has g = 1.63 m/s^2 and R = 1.74 x 10^6 m. Escape speed?
Check your answer
About 2.38 km/s.
sqrt(2 x 1.63 x 1.74 x 10^6) = sqrt(5.67 x 10^6) = 2380 m/s.
Orbital speed and period
Choose the method: v = sqrt(G M / r) and T = 2 pi r / v for a circular orbit of radius r.
EZT practice question
A satellite orbits at 400 km altitude, so r = 6800 km = 6.8 x 10^6 m, with G M = 4.00 x 10^14 m^3/s^2. Find its speed and period.
Show the worked solution
- v = sqrt(G M / r) = sqrt(4.00 x 10^14 / 6.8 x 10^6).
- T = 2 pi r / v.
Answer: v = 7.67 km/s, T = 92.4 minutes (approximately).
Check: Low-orbit satellites circle in about 90 minutes - the ISS is the familiar example.
Watch for this: Use the orbit radius r = R + h, not the altitude h alone, in both formulas.
Now try it yourself
A satellite at r = 4 R. Orbital speed relative to the surface value?
Check your answer
Half of sqrt(G M / R).
v scales as 1/sqrt(r), so quadrupling r halves the speed.
Use Kepler's third law
Choose the method: T^2 proportional to a^3 for orbits around the same central mass.
EZT practice question
Satellite A orbits at radius r; satellite B orbits at radius 4 r. How many times longer is B's period?
Show the worked solution
- T^2 / T_A^2 = (4r)^3 / r^3 = 64.
- T_B / T_A = sqrt(64).
Answer: B's period is 8 times A's.
Check: Periods grow faster than radii - far satellites take disproportionately longer, which is why geostationary orbits sit so far out.
Watch for this: T is proportional to a^1.5, not to a^3 - remember to take the square root.
Now try it yourself
B at radius 8 r. Period ratio?
Check your answer
About 22.6 times.
8^1.5 = sqrt(512) = 22.6.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Universal law?
Show answer and reason
F = G m1 m2 / r^2.
G = 6.67 x 10^-11 N m^2/kg^2.
2. g at height h?
Show answer and reason
g R^2 / (R + h)^2.
The 1 - 2h/R shortcut needs h much less than R.
3. Escape speed?
Show answer and reason
sqrt(2 g R), about 11.2 km/s for Earth.
Independent of the projectile's mass.
4. Orbital energy?
Show answer and reason
E = -G M m / 2r for a circular orbit.
Kinetic + half the potential cancels the other half.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Two masses of 10 kg and 20 kg with centres 2 m apart. Find the gravitational force.
- Find g at a height equal to Earth's radius.
- The Moon has g = 1.63 m/s^2 and R = 1.74 x 10^6 m. Estimate its escape speed.
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Questions about this practice page
- Why does g change with height?
- Because the inverse-square law spreads the pull over distance: g = G M / (R + h)^2 falls as you climb. The simple g(1 - 2h/R) formula is only an approximation for heights much smaller than Earth's radius.
- What exactly is escape speed?
- The minimum launch speed for an object to reach infinity without further propulsion - when its kinetic energy just cancels the (negative) gravitational potential energy. For Earth it is sqrt(2 g R), about 11.2 km/s.
- Does escape speed depend on the projectile's mass?
- No. The mass cancels between the kinetic energy and the potential energy, so everything from a pebble to a rocket shares the same escape speed from a given body.
- What are Kepler's three laws in one line each?
- Orbits are ellipses with the Sun at a focus; the radius line sweeps equal areas in equal times; and the period squared is proportional to the semi-major axis cubed.
- Are these the NCERT exercise questions for Gravitation (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.