Quick answer: Newton's law of gravitation says two masses attract with force F = G m₁m₂/r². This Physics 11-12 worksheet on gravitational field gives 20 problems from easy to hard: key relationships, two people close together and surface gravity of the Earth. The worked answers are on a separate page, and the worksheet is a free printable PDF.

These twenty problems cover Newton’s law of gravitation, field strength, potential, escape speed and circular orbits. You will use the inverse-square law, read a field graph, apply Kepler’s third law to planets and moons, and combine forces and energy for satellites and double stars.
They suit A Level, IB physics and CBSE Class 11 students. Use G = 6.67 × 10⁻¹¹ N m² kg⁻² and G M(Earth) = 3.98 × 10¹⁴ m³/s²; r is always measured from the centre of the planet.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB D.1
Problem 1 Key relationships ★★★
Write each relationship.1. Newton’s law of gravitation.
2. The field strength a distance r from a spherical mass M.
3. The gravitational potential at distance r from M.
4. The orbital speed and the period for a circular orbit.
Problem 2 Two people close together ★★★
Two students of mass 70 kg stand 1.0 m apart. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².1. Calculate the gravitational force between them.
2. Compare it with the weight of one student (use g = 9.81 N/kg).
Problem 3 Surface gravity of the Earth ★★★
The Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m.1. Calculate the field strength at the surface.
2. Explain why the answer is the same as the free-fall acceleration.
Problem 4 Weight on the Moon ★★★
The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. An astronaut has a mass of 60 kg.1. Calculate the field strength at the surface of the Moon.
2. Calculate the astronaut’s weight on the Moon and on the Earth.
3. State the astronaut’s mass on the Moon.
Problem 5 Inverse-square thinking ★★★
The gravitational force between two masses is F. State the new force when:1. the distance is tripled;
2. one mass is doubled;
3. both masses are doubled and the distance is doubled;
4. the distance is halved.
Problem 6 Potential at the surface ★★★
Use G M = 3.98 × 10¹⁴ m³/s² and R = 6.37 × 10⁶ m for the Earth.1. Calculate the potential at the Earth’s surface.
2. State what the negative sign means.
3. Calculate the energy needed to take a 2.0 kg mass from the surface to infinity.
Problem 7 Moons of Jupiter ★★★
Io orbits Jupiter in a nearly circular orbit of radius 4.22 × 10⁸ m with a period of 1.77 days.1. Convert the period to seconds.
2. Use T² = 4π² r³ ÷ (G M) to find the mass of Jupiter.
Problem 8 Half the surface gravity ★★★
Take the Earth’s radius as R = 6.37 × 10⁶ m.1. Show that g is half its surface value when r = √2 R.
2. Calculate the height above the surface at which g = 4.9 N/kg.
Problem 9 Reading a g–r graph ★★★
The graph shows how the Earth’s field strength varies with distance from its centre.2. Show that g r² is constant for the points at R, 2R and 3R.
3. Calculate the change in potential between R and 3R.
Problem 10 Orbit of the space station ★★★
A space station orbits 410 km above the Earth’s surface. Use G M = 3.98 × 10¹⁴ m³/s² and R = 6.37 × 10⁶ m.1. Calculate the orbital radius.
2. Calculate the orbital speed.
3. Calculate the period in minutes.
Problem 11 Kepler on the planets ★★★
Use T² ∝ r³ with T in years and r in astronomical units (AU), so that the Earth has T = 1 year and r = 1 AU.1. Mars orbits at 1.52 AU. Calculate its period.
2. A comet has a period of 76 years. Calculate the mean orbital radius (the semi-major axis) in AU.
Problem 12 Weighing the Sun ★★★
The Earth orbits the Sun in an almost circular orbit of radius 1.50 × 10¹¹ m with a period of 3.156 × 10⁷ s (one year).1. Show that the mass of the Sun is M = 4π² r³ ÷ (G T²).
2. Calculate the mass of the Sun.
3. Compare it with the Earth’s mass of 5.97 × 10²⁴ kg.
Problem 13 Leaving the Earth and Moon ★★★
Use G M = 3.98 × 10¹⁴ m³/s² and R = 6.37 × 10⁶ m for the Earth; the Moon has G M = 4.90 × 10¹² m³/s² and R = 1.74 × 10⁶ m.1. Calculate the escape speed from the Earth.
2. Calculate the escape speed from the Moon.
3. Calculate the kinetic energy a 500 kg probe needs at the Earth’s surface to escape.
Problem 14 Sun against Moon ★★★
The Sun has mass 1.99 × 10³⁰ kg and is 1.50 × 10¹¹ m from the Earth. The Moon has mass 7.35 × 10²² kg and is 3.84 × 10⁸ m from the Earth.1. Calculate the ratio of the Sun’s pull on the Earth to the Moon’s pull.
2. Suggest why the Moon still causes larger tides than the Sun.
Problem 15 No gravity in orbit? ★★★
A student says: “Astronauts on the space station float because there is no gravity at 400 km up.”1. Calculate the field strength at 400 km above the surface (r = 6.77 × 10⁶ m).
2. Express it as a percentage of 9.81 N/kg.
3. Explain correctly why astronauts appear weightless.
Problem 16 The neutral point ★★★
At a point P on the line between the Earth and the Moon the fields cancel. Use ME = 5.97 × 10²⁴ kg, MM = 7.35 × 10²² kg and d = 3.84 × 10⁸ m.2. Show that x ÷ (d − x) = √(ME ÷ MM).
3. Calculate x.
Problem 17 Energy to reach orbit ★★★
A satellite of mass 1200 kg is lifted from the surface of the Earth to a circular orbit of radius 8.0 × 10⁶ m. Ignore the rotation of the Earth and use G M = 3.98 × 10¹⁴ m³/s², R = 6.37 × 10⁶ m.1. Calculate the kinetic energy in the orbit.
2. Calculate the total energy in the orbit.
3. Calculate the total energy on the ground.
4. Calculate the minimum energy that must be supplied.
Problem 18 Binary stars ★★★
Two stars, each of mass 2.0 × 10³⁰ kg, orbit their common centre with a separation of 3.0 × 10¹¹ m.2. Show that ω² = 2 G M ÷ d³.
3. Calculate the period in years.
4. Calculate the orbital speed of each star.
Problem 19 Orbit decay paradox ★★★
A satellite of mass 500 kg orbits at r₁ = 6.78 × 10⁶ m. Atmospheric drag lowers it to a new circular orbit of radius r₂ = 6.68 × 10⁶ m. Use G M = 3.98 × 10¹⁴ m³/s².1. Calculate the orbital speed at r₁ and at r₂.
2. Calculate the total energy at each radius.
3. Explain why the satellite speeds up although drag acts against its motion.
Problem 20 Atmosphere of the Moon ★★★
The root-mean-square speed of gas molecules is v = √(3 k T ÷ m), with k = 1.38 × 10⁻²³ J/K. At 300 K a nitrogen molecule has m = 4.65 × 10⁻²⁶ kg and a hydrogen molecule has m = 3.34 × 10⁻²⁷ kg. A gas escapes over geological time if its mean speed exceeds about one sixth of the escape speed.1. Calculate the r.m.s. speed of nitrogen and of hydrogen at 300 K.
2. Calculate one sixth of the escape speed for the Earth (11.2 km/s) and for the Moon (2.37 km/s).
3. Predict which gases the Earth and the Moon retain.

