Cambridge AS & A Level9702

Gravitational fields

Physics 9702 Chapter Notes

What this chapter covers

Gravitational fields - Gravitational fieldGravitational fields - Gravitational force between point massesGravitational fields - Gravitational field of a point massGravitational fields - Gravitational potential
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In this chapter

  1. 1Understanding Gravitational Fields
  2. 2Newton's Law of Universal Gravitation
  3. 3Gravitational Field Strength (g)
  4. 4Gravitational Potential
  5. 5Satellite Orbits
  6. 6Geostationary Orbits

1. Understanding Gravitational Fields

A gravitational field is a region of space where an object with mass experiences a force. Think of it as an invisible influence created by any object that has mass, like a planet, star, or even you. When another mass enters this field, it gets 'told' how to move – it's pulled towards the object that created the field. We can visualize these fields using 'field lines'. These lines show the path a small test mass would take if placed in the field. For a single mass like a planet, the field is radial, with lines pointing straight to the centre. The closer the field lines are to each other, the stronger the gravitational force. Very close to a massive object like the Earth, the field lines are nearly parallel and evenly spaced, so we call the field 'uniform'. In this uniform field, the gravitational pull is considered constant.

Key term

Gravitational Field: A region of space where a mass experiences a non-contact force due to the presence of another mass.

Examiner insight

Examiners look for clearly drawn field line diagrams. Lines must be radial for a planet, have arrows pointing inwards, and should not cross each other.

Worked example 13 marks

Sketch the gravitational field around a uniform spherical planet. On your diagram, label a point A close to the surface and a point B further away. Compare the field strength at A and B.

  1. 1

    Step 1: Draw a circle to represent the planet.

  2. 2

    Step 2: Draw at least 6-8 straight lines starting from outside the circle and pointing directly towards the centre. These are the field lines. Add arrows to each line to show the direction of the force is attractive (inwards).

  3. 3

    Step 3: Ensure the lines are further apart as they get further from the planet's surface. This represents the field getting weaker with distance.

  4. 4

    Step 4: Label a point A close to the surface where the field lines are relatively close together. Label a point B further out where the lines are more spread out.

  5. 5

    Step 5: State that the field strength at A is greater than the field strength at B because the field lines at A are more densely packed.

Recap

  • A gravitational field is a region where a mass experiences a force.
  • Field lines show the direction of the gravitational force (always attractive).
  • The spacing of field lines indicates the strength of the field; closer lines mean a stronger field.
  • A field around a planet is radial, while a field over a small area on its surface is considered uniform.

Quick check

  1. What two things do gravitational field lines tell us about the field?2 marks
  2. Why is the gravitational field in a classroom described as uniform?1 mark

2. Newton's Law of Universal Gravitation

Sir Isaac Newton figured out the rule that governs the force of gravity everywhere in the universe. His law states that any two objects with mass attract each other with a force. This force is stronger if the masses are larger (it's directly proportional to the product of their masses, m₁m₂) and gets much weaker as they get further apart (it's inversely proportional to the square of the distance between their centres, r²). This 'inverse square law' is a very important concept. Doubling the distance reduces the force to a quarter of its original value. The full equation includes a special number, 'G', the universal gravitational constant, which is the same everywhere in the universe.

F = G * m₁ * m₂ / r²

Key term

Newton's Law of Gravitation: The attractive force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation.

Common pitfall

A common mistake is forgetting that 'r' is the distance between the centres of the two masses, not from the surface of one to the other.

Fun fact

Henry Cavendish was the first person to accurately measure the value of G in 1798. His experiment was so sensitive it was described as 'weighing the Earth'.

Worked example 13 marks

Calculate the magnitude of the gravitational force between the Earth and the Moon. (Mass of Earth = 5.97 x 10²⁴ kg, Mass of Moon = 7.35 x 10²² kg, Average separation = 3.84 x 10⁸ m, G = 6.67 x 10⁻¹¹ N m² kg⁻²).

  1. 1

    Step 1: Write down Newton's Law of Gravitation: F = G * M * m / r².

  2. 2

    Step 2: Substitute the given values into the equation.

  3. 3

    F = (6.67 x 10⁻¹¹) * (5.97 x 10²⁴) * (7.35 x 10²²) / (3.84 x 10⁸)²

  4. 4

    Step 3: Calculate the numerator: F = (2.925 x 10³⁷) / (3.84 x 10⁸)²

  5. 5

    Step 4: Calculate the denominator: F = (2.925 x 10³⁷) / (1.475 x 10¹⁷)

  6. 6

    Step 5: Calculate the final force: F = 1.98 x 10²⁰ N.

Recap

  • Gravitational force is always attractive.
  • The force is proportional to the product of the two masses.
  • The force follows an inverse square law with the distance between the centres of the masses.
  • G is the universal gravitational constant, a value you will be given in exams.

Quick check

  1. If the distance between two planets is halved, by what factor does the gravitational force between them change?2 marks

3. Gravitational Field Strength (g)

Gravitational field strength, symbol 'g', is a measure of how strong a gravitational field is at a certain point. It's defined as the force experienced per unit mass. So, if you place a 1 kg mass at a point, the force it feels in Newtons is numerically equal to the gravitational field strength 'g' at that point. Its unit is Newtons per kilogram (N kg⁻¹), which is equivalent to the familiar unit for acceleration, m s⁻². On Earth's surface, g ≈ 9.81 N kg⁻¹. By combining the definition g = F/m with Newton's Law F = GMm/r², we can derive a more useful formula for the field strength caused by a large mass M: g = GM/r². This shows that the field strength depends on the mass of the object creating the field and your distance from its centre.

g = F / m

g = G * M / r²

Key term

Gravitational Field Strength (g): The gravitational force exerted per unit mass on a small test mass placed at that point in the field.

Examiner insight

Showing the derivation of g = GM/r² from first principles (F=GMm/r² and g=F/m) is a good way to demonstrate a thorough understanding of the concepts.

Fun fact

Your weight is slightly less on top of a tall mountain than at sea level, because 'g' decreases as your distance 'r' from the Earth's centre increases.

Worked example 13 marks

The mass of Mars is 6.42 x 10²³ kg and its radius is 3.39 x 10⁶ m. Calculate the gravitational field strength on the surface of Mars. (G = 6.67 x 10⁻¹¹ N m² kg⁻²)

  1. 1

    Step 1: State the formula for gravitational field strength: g = GM/r².

  2. 2

    Step 2: Identify the values. M = 6.42 x 10²³ kg and r = 3.39 x 10⁶ m.

  3. 3

    Step 3: Substitute the values into the formula: g = (6.67 x 10⁻¹¹) * (6.42 x 10²³) / (3.39 x 10⁶)².

  4. 4

    Step 4: Calculate the result: g = (4.282 x 10¹³) / (1.149 x 10¹³) = 3.73 N kg⁻¹.

Worked example 24 marks

At what altitude above the Earth's surface is the gravitational field strength 25% of its value at the surface? (Radius of Earth, Rₑ = 6.4 x 10⁶ m)

  1. 1

    Step 1: Let gₛ be the field strength at the surface (r = Rₑ) and gₐ be the field strength at altitude h (r = Rₑ + h). gₛ = GM/Rₑ².

  2. 2

    Step 2: The condition is gₐ = 0.25 * gₛ. The formula for gₐ is gₐ = GM / (Rₑ + h)².

  3. 3

    Step 3: Set up the equation: GM / (Rₑ + h)² = 0.25 * (GM / Rₑ²).

  4. 4

    Step 4: Cancel GM from both sides: 1 / (Rₑ + h)² = 0.25 / Rₑ². Rearrange to get (Rₑ + h)² = Rₑ² / 0.25 = 4Rₑ².

  5. 5

    Step 5: Take the square root of both sides: Rₑ + h = 2Rₑ.

  6. 6

    Step 6: Solve for h: h = 2Rₑ - Rₑ = Rₑ. The altitude is equal to the Earth's radius, h = 6.4 x 10⁶ m.

Recap

  • Gravitational field strength 'g' is the force per unit mass at a point.
  • The unit of g is N kg⁻¹, which is equivalent to m s⁻².
  • g is constant in a uniform field but varies as 1/r² in a radial field.
  • You can derive g = GM/r² from Newton's Law and the definition of g.

Quick check

  1. Distinguish between the terms 'g' and 'G'.2 marks

4. Gravitational Potential

Gravitational potential (symbol Φ, the Greek letter phi) is a concept related to energy. The gravitational potential at a point in a field is defined as the work done per unit mass to move an object from infinity to that point. Since gravity is an attractive force, the field does positive work on the mass as it comes from infinity, meaning we don't have to put in energy. Therefore, the potential energy of the mass decreases. By convention, we say the potential at infinity is zero. As a mass moves from infinity (where potential is zero) closer to a planet, its potential becomes negative and gets smaller (more negative). The formula is Φ = -GM/r. The negative sign is very important! Gravitational Potential Energy (GPE) of a mass 'm' is simply its mass multiplied by the potential at that point: GPE = mΦ, which gives the full formula GPE = -GMm/r.

Φ = -G * M / r

GPE = -G * M * m / r

Work Done = m * ΔΦ = m * (Φ_final - Φ_initial)

Key term

Gravitational Potential (Φ): The work done per unit mass in bringing a small test mass from infinity to a point in a gravitational field.

Common pitfall

Forgetting the negative sign in the potential and potential energy equations. This is a critical feature that shows energy must be added to an object to move it to infinity (to escape the field).

Worked example 13 marks

Calculate the gravitational potential at the surface of the Earth. (Mass of Earth = 5.97 x 10²⁴ kg, Radius of Earth = 6.37 x 10⁶ m, G = 6.67 x 10⁻¹¹ N m² kg⁻²).

  1. 1

    Step 1: Write down the formula for gravitational potential: Φ = -GM/r.

  2. 2

    Step 2: Substitute the values for Earth's mass and radius: Φ = -(6.67 x 10⁻¹¹) * (5.97 x 10²⁴) / (6.37 x 10⁶).

  3. 3

    Step 3: Calculate the numerator: Φ = -(3.982 x 10¹⁴) / (6.37 x 10⁶).

  4. 4

    Step 4: Calculate the final value: Φ = -6.25 x 10⁷ J kg⁻¹.

Worked example 24 marks

Calculate the work done against gravity to lift a 1200 kg satellite from the Earth's surface to an altitude of 2000 km. Use the potential value from the previous example.

  1. 1

    Step 1: First, calculate the potential at the new altitude. The new radius is r_final = 6.37 x 10⁶ m + 2000 x 10³ m = 8.37 x 10⁶ m.

  2. 2

    Step 2: Φ_final = -GM/r_final = -(6.67 x 10⁻¹¹) * (5.97 x 10²⁴) / (8.37 x 10⁶) = -4.76 x 10⁷ J kg⁻¹.

  3. 3

    Step 3: The initial potential at the surface is Φ_initial = -6.25 x 10⁷ J kg⁻¹.

  4. 4

    Step 4: The change in potential is ΔΦ = Φ_final - Φ_initial = (-4.76 x 10⁷) - (-6.25 x 10⁷) = +1.49 x 10⁷ J kg⁻¹.

  5. 5

    Step 5: Work done = mass * ΔΦ = 1200 kg * (1.49 x 10⁷ J kg⁻¹) = 1.79 x 10¹⁰ J.

Recap

  • Gravitational potential at infinity is defined as zero.
  • Since gravity is attractive, gravitational potential is always negative.
  • Potential becomes more negative as you get closer to the mass creating the field.
  • Gravitational potential is a scalar quantity, measured in J kg⁻¹.
  • Work done in moving a mass is the mass multiplied by the change in potential.

Quick check

  1. What is the key difference between gravitational potential and gravitational potential energy?2 marks
  2. Why must gravitational potential always be a negative value?1 mark

5. Satellite Orbits

For a satellite to maintain a stable circular orbit around a planet, there must be a force constantly pulling it towards the centre of the circle. This force is called the centripetal force. In the case of an orbiting satellite, the planet's gravitational force provides this centripetal force. By setting the formula for gravitational force equal to the formula for centripetal force (F = mv²/r), we can analyse the motion. The equation becomes GMm/r² = mv²/r. Notice that the mass of the satellite, 'm', cancels out! This means the speed required for an orbit at a certain radius 'r' is the same for any satellite, whether it's a tiny CubeSat or the massive International Space Station. From this relationship, we can derive formulas for the satellite's orbital speed 'v' and its orbital period 'T' (the time for one full orbit).

G * M * m / r² = m * v² / r

v² = G * M / r

T² = (4π² / (G * M)) * r³

Key term

Centripetal Force: The resultant force acting on an object moving in a circle, which is directed towards the centre of the circle and is required to maintain the circular motion.

Common pitfall

Treating the gravitational force and centripetal force as two separate forces in a free-body diagram. The gravitational force *is* the centripetal force in this situation.

Fun fact

Astronauts on the ISS are not floating because there is 'no gravity'. At their altitude, gravity is about 90% as strong as on the surface. They feel weightless because they, and the station around them, are in a constant state of freefall.

Worked example 14 marks

The International Space Station (ISS) orbits at an average altitude of 408 km above the Earth's surface. Calculate its orbital speed. (Mass of Earth = 5.97 x 10²⁴ kg, Radius of Earth = 6.37 x 10⁶ m, G = 6.67 x 10⁻¹¹ N m² kg⁻²).

  1. 1

    Step 1: First, calculate the total orbital radius, r. This is the Earth's radius plus the altitude. r = (6.37 x 10⁶m) + (408 x 10³m) = 6.778 x 10⁶ m.

  2. 2

    Step 2: State the relationship for orbital speed: v² = GM/r.

  3. 3

    Step 3: Substitute the known values: v² = (6.67 x 10⁻¹¹) * (5.97 x 10²⁴) / (6.778 x 10⁶).

  4. 4

    Step 4: Calculate v²: v² = (3.982 x 10¹⁴) / (6.778 x 10⁶) = 5.875 x 10⁷ m²s⁻².

  5. 5

    Step 5: Take the square root to find v: v = √(5.875 x 10⁷) = 7665 m s⁻¹. (or ~7.7 km/s)

Recap

  • For a stable orbit, the gravitational force provides the centripetal force.
  • The mass of the satellite does not affect its orbital speed or period.
  • A satellite in a higher orbit (larger r) moves more slowly than one in a lower orbit.
  • The relationship T² ∝ r³ (Kepler's Third Law) applies to all satellites orbiting the same central mass.

Quick check

  1. What force provides the centripetal force for the Moon's orbit around the Earth?1 mark
  2. Two satellites of different masses orbit a planet at the same radius. Which one has a shorter orbital period?1 mark

6. Geostationary Orbits

A geostationary orbit is a special, very useful type of orbit. A satellite in this orbit appears to hover over the same spot on the Earth's surface. For this to happen, three conditions must be met: 1. The satellite's orbital period must be exactly 24 hours (to be precise, one sidereal day: 23 hours, 56 minutes, 4 seconds), so it matches the Earth's rotation. 2. The orbit must be directly above the equator. If it were inclined, it would appear to move north and south in the sky. 3. The satellite must orbit in the same direction as the Earth's rotation (from west to east). Because the satellite stays in a fixed position relative to the ground, it is ideal for telecommunications and weather broadcasting, as a fixed satellite dish on the ground can always point to it.

T² = (4π² / (G * M)) * r³ (where T = 24 hours)

Key term

Geostationary Orbit: A circular orbit in the Earth's equatorial plane with a period of 24 hours, meaning the satellite remains above the same point on the Earth's surface.

Examiner insight

Questions asking for the conditions of a geostationary orbit are very common. Memorising the three key conditions (period, plane, direction) is essential for securing all the marks.

Worked example 15 marks

Calculate the radius of a geostationary orbit around the Earth. (Mass of Earth = 5.97 x 10²⁴ kg, G = 6.67 x 10⁻¹¹ N m² kg⁻²).

  1. 1

    Step 1: State the formula relating period and radius: T² = (4π² / GM) * r³.

  2. 2

    Step 2: The period T must be 24 hours in seconds. T = 24 hours * 60 min/hr * 60 s/min = 86400 s.

  3. 3

    Step 3: Rearrange the formula to solve for r³: r³ = G * M * T² / (4π²).

  4. 4

    Step 4: Substitute the values: r³ = (6.67 x 10⁻¹¹) * (5.97 x 10²⁴) * (86400)² / (4π²).

  5. 5

    Step 5: Calculate the numerator and denominator: r³ = (3.982 x 10¹⁴) * (7.465 x 10⁹) / 39.48 = 2.97 x 10²⁴ / 39.48 = 7.52 x 10²² m³.

  6. 6

    Step 6: Take the cube root to find r: r = ³√(7.52 x 10²²) = 4.22 x 10⁷ m. This is the radius from the centre of the Earth.

Recap

  • A geostationary satellite appears fixed in the sky from the Earth.
  • It must have an orbital period of 24 hours.
  • It must orbit in the equatorial plane.
  • It must travel in the same direction as the Earth's rotation (west to east).
  • All geostationary satellites orbit at the same specific altitude.

Quick check

  1. State two applications of geostationary satellites.2 marks
  2. Why can't a geostationary satellite orbit over London?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define gravitational field strength and state its standard unit.2 marks
  2. The planet Jupiter has a mass of 1.90 x 10²⁷ kg and a radius of 6.99 x 10⁷ m. An uncrewed probe of mass 800 kg is in a circular orbit at an altitude of 2.00 x 10⁸ m above Jupiter's surface. Calculate the magnitude of the gravitational force on the probe. (G = 6.67 x 10⁻¹¹ N m² kg⁻²)4 marks
  3. Derive the formula T² = (4π²/GM)r³ for a satellite in a circular orbit, starting from the fact that the gravitational force provides the centripetal force.4 marks
  4. Explain the three essential conditions that a satellite must satisfy to be in a geostationary orbit relative to the Earth.3 marks
  5. Calculate the change in gravitational potential energy of a 1500 kg satellite when it is moved from a stable orbit at an altitude of 500 km to one at 1000 km. (Mass of Earth = 5.97 x 10²⁴ kg, Radius of Earth = 6.37 x 10⁶ m, G = 6.67 x 10⁻¹¹ N m² kg⁻²)5 marks
  6. Sketch the gravitational field pattern produced by the Earth and Moon system. Your diagram should show the field lines and indicate the neutral point where the net gravitational field strength is zero.4 marks
  7. A probe is sent from Earth towards the Sun. The mass of the Sun is 3.3 x 10⁵ times the mass of the Earth. The distance between the centres of the Earth and Sun is 1.5 x 10¹¹ m. Calculate the distance from the Earth's centre where the net gravitational field strength due to the Earth and Sun is zero.5 marks
  8. Explain what is meant by the term 'gravitational potential' at a point, and explain why it is always a negative value.3 marks
  9. A satellite is in a circular orbit around a planet of mass M. Show that the total energy of the satellite (kinetic + potential) is given by E = -GMm / 2r.6 marks
  10. A spy satellite is in a low-Earth orbit with a period of 90 minutes. Calculate its altitude above the Earth's surface. (Mass of Earth = 5.97 x 10²⁴ kg, Radius of Earth = 6.37 x 10⁶ m, G = 6.67 x 10⁻¹¹ N m² kg⁻²)4 marks

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