IB Diploma ProgrammeHL

Gravitational fields

Physics HL Chapter Notes

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Gravitational fieldsElectric and magnetic fieldsMotion in electromagnetic fieldsInduction
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1. Newton's Law of Universal Gravitation

At the heart of celestial mechanics is a simple but profound idea from Isaac Newton: every object with mass pulls on every other object with mass. This pull is called the gravitational force. Newton's Law of Universal Gravitation states that this force is directly proportional to the product of the two masses (the more massive the objects, the stronger the pull) and inversely proportional to the square of the distance between their centers (the force gets weaker very quickly as the objects move apart). This is known as an 'inverse square law'. The force acts along the line connecting the centers of the two objects. For spherical objects like planets and stars, we can treat all their mass as being concentrated at a single point at their center.

F = G * (M * m) / r^2

Key term

Newton's Law of Universal Gravitation: The law states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Examiner insight

Examiners look for the correct substitution of values into the formula and a clear final answer given to an appropriate number of significant figures.

Common pitfall

Forgetting that 'r' is the distance between the centers of the masses, not their surfaces. For an object on a planet's surface, 'r' is the planet's radius.

Fun fact

The gravitational force is the weakest of the four fundamental forces of nature, yet it governs the large-scale structure of the entire universe.

Worked example 13 marks

The planet Mars has a mass of 6.42 x 10^23 kg and a radius of 3.39 x 10^6 m. Its moon, Phobos, has a mass of 1.06 x 10^16 kg and orbits at a distance of 9.38 x 10^6 m from the center of Mars. Calculate the magnitude of the gravitational force between Mars and Phobos. (G = 6.67 x 10^-11 N m^2 kg^-2)

  1. 1
    1. Identify the relevant formula: F = G * (M * m) / r^2.
  2. 2
    1. List the given values: G = 6.67 x 10^-11, M (Mars) = 6.42 x 10^23 kg, m (Phobos) = 1.06 x 10^16 kg, r = 9.38 x 10^6 m.
  3. 3
    1. Substitute the values into the formula: F = (6.67 x 10^-11) * (6.42 x 10^23 * 1.06 x 10^16) / (9.38 x 10^6)^2.
  4. 4
    1. Calculate the product of the masses: (6.42 x 10^23) * (1.06 x 10^16) = 6.8052 x 10^39 kg^2.
  5. 5
    1. Calculate the square of the distance: (9.38 x 10^6)^2 = 8.79844 x 10^13 m^2.
  6. 6
    1. Complete the calculation: F = (6.67 x 10^-11) * (6.8052 x 10^39) / (8.79844 x 10^13) = 5.16 x 10^15 N.

Recap

  • Gravitational force is always attractive.
  • The force is proportional to the product of the masses (M*m).
  • The force is inversely proportional to the square of the distance between centers (1/r^2).
  • G is the Universal Gravitational Constant, approximately 6.67 x 10^-11 N m^2 kg^-2.
  • For spherical bodies, 'r' is the distance between their centers.

Quick check

  1. If the distance between two point masses is halved, by what factor does the gravitational force between them change?2 marks
  2. What does the 'G' in Newton's law represent?1 mark

2. Gravitational Field Strength (g)

A mass creates a gravitational field in the space around it. This field is a region where another mass will experience a gravitational force. We quantify the strength of this field using the concept of gravitational field strength, symbol 'g'. It is defined as the gravitational force experienced per unit mass at a particular point. So, g = F/m. Since force is a vector, field strength is also a vector, and its direction is the same as the direction of the force – always towards the mass creating the field. By combining this definition with Newton's Law of Gravitation, we can derive a formula for the field strength created by a point or spherical mass M: g = GM/r^2. The units of 'g' are N kg^-1, which are equivalent to m s^-2, meaning gravitational field strength is numerically equal to the acceleration of free fall at that point.

g = F / m

g = G * M / r^2

Key term

Gravitational Field Strength (g): The gravitational force experienced per unit mass at a point in a gravitational field.

Examiner insight

Examiners often award marks for correctly stating that gravitational field strength is a vector quantity and specifying its direction (e.g., 'radially inwards' or 'towards the Earth's center').

Common pitfall

When calculating field strength at an altitude, students often use the altitude itself as 'r' instead of adding it to the planet's radius.

Worked example 13 marks

The Earth has a mass of 5.97 x 10^24 kg and a mean radius of 6.37 x 10^6 m. Calculate the gravitational field strength on the surface of the Earth.

  1. 1
    1. The formula for gravitational field strength at the surface of a spherical mass M is g = GM/r^2.
  2. 2
    1. Identify the values: G = 6.67 x 10^-11 N m^2 kg^-2, M = 5.97 x 10^24 kg, and r = 6.37 x 10^6 m.
  3. 3
    1. Substitute the values: g = (6.67 x 10^-11 * 5.97 x 10^24) / (6.37 x 10^6)^2.
  4. 4
    1. Calculate the numerator: (6.67 x 10^-11) * (5.97 x 10^24) = 3.982 x 10^14.
  5. 5
    1. Calculate the denominator: (6.37 x 10^6)^2 = 4.058 x 10^13.
  6. 6
    1. Divide to find g: g = (3.982 x 10^14) / (4.058 x 10^13) = 9.81 N kg^-1.

Worked example 24 marks

A satellite orbits at an altitude of 400 km above the Earth's surface. Using the data from the previous example, calculate the gravitational field strength at this altitude.

  1. 1
    1. First, determine the total distance 'r' from the center of the Earth. This is the Earth's radius plus the altitude.
  2. 2
    1. Convert altitude to meters: 400 km = 400 x 10^3 m.
  3. 3
    1. Calculate 'r': r = (6.37 x 10^6 m) + (400 x 10^3 m) = 6.77 x 10^6 m.
  4. 4
    1. Use the formula g = GM/r^2 with the new value of 'r'.
  5. 5
    1. Substitute values: g = (6.67 x 10^-11 * 5.97 x 10^24) / (6.77 x 10^6)^2.
  6. 6
    1. Calculate the result: g = (3.982 x 10^14) / (4.583 x 10^13) = 8.69 N kg^-1.

Recap

  • Gravitational field strength 'g' is the force per unit mass (F/m).
  • For a spherical mass M, the field strength is given by g = GM/r^2.
  • 'g' is a vector quantity directed towards the center of the mass M.
  • The units N kg^-1 and m s^-2 are equivalent.
  • Field strength decreases with the square of the distance from the center.

Quick check

  1. A planet has a surface gravitational field strength of 'g'. What is the field strength at a height of one planet-radius above the surface?2 marks

3. Superposition of Gravitational Fields

What happens when you are influenced by more than one gravitational field, for example, at a point between the Earth and the Moon? The Principle of Superposition provides the answer. It states that the total gravitational field strength at any point is the vector sum of the individual field strengths from all masses present. If the fields from two masses point in the same direction, you add their magnitudes. If they point in opposite directions (like at a point on the line between two planets), you subtract the smaller magnitude from the larger one. This principle allows us to find points in space where the net gravitational field is zero. Such a point is called a 'null point' or 'neutral point'.

g_total = g_1 + g_2 + ... (vector sum)

Key term

Principle of Superposition: For a system of masses, the net gravitational field strength at any point is the vector sum of the gravitational field strengths due to each individual mass.

Examiner insight

In problems involving a null point, examiners look for a clear statement that the magnitudes of the two fields are equal at that point, and a correctly set-up equation.

Common pitfall

Treating gravitational field strength as a scalar and always adding the magnitudes, instead of performing a vector addition which often involves subtraction for points between two masses.

Worked example 15 marks

Two point masses, M1 = 4.0 x 10^24 kg and M2 = 1.0 x 10^24 kg, are separated by a distance of 3.0 x 10^8 m. Determine the point P on the line between the two masses where the net gravitational field strength is zero.

  1. 1
    1. Let the point P be a distance 'x' from the larger mass M1. Its distance from M2 will therefore be (3.0 x 10^8 - x).
  2. 2

    2. At point P, the gravitational field g1 due to M1 is directed towards M1, and the field g2 due to M2 is directed towards M2. For the net field to be zero, the magnitudes must be equal: |g1| = |g2|.

  3. 3
    1. Write the equation using the formula for g: G*M1/x^2 = G*M2/(3.0 x 10^8 - x)^2.
  4. 4
    1. The constant G cancels from both sides: M1/x^2 = M2/(3.0 x 10^8 - x)^2.
  5. 5
    1. Substitute the mass values: (4.0 x 10^24)/x^2 = (1.0 x 10^24)/(3.0 x 10^8 - x)^2.
  6. 6
    1. Simplify by dividing by 1.0 x 10^24: 4/x^2 = 1/(3.0 x 10^8 - x)^2.
  7. 7
    1. Take the square root of both sides: 2/x = 1/(3.0 x 10^8 - x).
  8. 8
    1. Rearrange to solve for x: 2 * (3.0 x 10^8 - x) = x => 6.0 x 10^8 - 2x = x.
  9. 9
    1. Solve for x: 3x = 6.0 x 10^8 => x = 2.0 x 10^8 m.
  10. 10
    1. State the final answer: The point is 2.0 x 10^8 m from the larger mass (M1).

Recap

  • The net gravitational field is the vector sum of individual fields.
  • For fields in opposite directions, subtract their magnitudes.
  • A null point occurs where the magnitudes of the opposing fields are equal.
  • The null point between two masses is always closer to the smaller mass.

Quick check

  1. Two planets have equal mass. At the midpoint between their centers, what is the magnitude of the net gravitational field strength?1 mark

4. Gravitational Potential and Energy

To understand energy in a gravitational field, we must first define a zero point. In physics, we define the gravitational potential energy (GPE) of a mass as zero when it is at an infinite distance from the source mass. To bring the mass from infinity to a point 'r' in the field, the gravitational force does work on the mass. Because the force is attractive, the mass 'falls' towards the source, and its GPE decreases from zero. Therefore, the gravitational potential energy of any mass in a gravitational field is always negative. We also define a related quantity, gravitational potential (V_g), as the gravitational potential energy per unit mass (V_g = E_p / m). It represents the work done per unit mass to bring a test mass from infinity to that point. Like GPE, gravitational potential is also always negative and is a scalar quantity, which makes energy calculations much simpler than vector force calculations.

V_g = -G * M / r

E_p = -G * M * m / r

Work Done = m * ΔV_g

Key term

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

Examiner insight

Students who can clearly explain why gravitational potential is always negative (by defining the zero point at infinity and noting that the force is attractive) often gain extra credit in descriptive questions.

Common pitfall

Forgetting the negative sign in the formulas for potential and potential energy. This is a critical error as the sign carries important physical meaning about the system being bound.

Worked example 13 marks

Calculate the gravitational potential at the surface of the Moon. The Moon's mass is 7.35 x 10^22 kg and its radius is 1.74 x 10^6 m.

  1. 1
    1. The formula for gravitational potential is V_g = -GM/r.
  2. 2
    1. List the values: G = 6.67 x 10^-11, M = 7.35 x 10^22 kg, r = 1.74 x 10^6 m.
  3. 3
    1. Substitute the values: V_g = -(6.67 x 10^-11 * 7.35 x 10^22) / (1.74 x 10^6).
  4. 4
    1. Calculate the numerator: -(4.90245 x 10^12).
  5. 5
    1. Divide to find V_g: V_g = -(4.90245 x 10^12) / (1.74 x 10^6) = -2.82 x 10^6 J kg^-1.

Worked example 23 marks

Using the result from the previous example, find the work required to lift a 1500 kg spacecraft from the surface of the Moon to a very large distance away (effectively infinity).

  1. 1
    1. The work done is equal to the change in gravitational potential energy: W = ΔE_p.
  2. 2
    1. ΔE_p = E_p(final) - E_p(initial).
  3. 3
    1. The final GPE at infinity is 0. The initial GPE on the surface is E_p = m * V_g.
  4. 4
    1. Calculate initial GPE: E_p(initial) = 1500 kg * (-2.82 x 10^6 J kg^-1) = -4.23 x 10^9 J.
  5. 5
    1. Calculate the work done: W = 0 - (-4.23 x 10^9 J) = +4.23 x 10^9 J.
  6. 6
    1. The positive sign indicates that energy must be supplied to the spacecraft to move it away from the Moon.

Recap

  • Gravitational potential (V_g) is the GPE per unit mass.
  • Gravitational potential energy (E_p) and potential (V_g) are zero at infinity.
  • Both E_p and V_g are always negative for any finite distance 'r'.
  • The negative sign indicates an attractive force and a bound system.
  • Potential and potential energy are scalar quantities, so you just add their values.
  • The work done to move a mass 'm' is m multiplied by the change in potential (mΔV_g).

Quick check

  1. Why is gravitational potential energy always a negative value?2 marks
  2. What is the unit of gravitational potential?1 mark

5. Equipotential Surfaces

An equipotential surface is a surface where the gravitational potential is the same at every point. For a single spherical mass, these surfaces are concentric spheres. Since potential is constant everywhere on an equipotential, the change in potential (ΔV_g) for any movement along the surface is zero. As the work done is mΔV_g, this means no work is done when moving a mass along an equipotential surface. There is a direct link between field strength and potential: the gravitational field strength 'g' is equal to the negative of the potential gradient. This means g = -ΔV_g/Δr. In simpler terms, the field is strongest where the equipotential surfaces are closest together. A key property is that gravitational field lines are always perpendicular to equipotential surfaces and point in the direction of decreasing potential (i.e., towards more negative values).

g ≈ -ΔV_g / Δr

Key term

Equipotential Surface: A surface consisting of points that all have the same gravitational potential.

Examiner insight

Marks are often available for drawing field lines and equipotentials correctly, specifically showing that field lines are perpendicular to equipotential surfaces and are more densely spaced where the field is stronger.

Worked example 14 marks

The diagram shows two equipotential surfaces around a planet. Surface A has a potential of -40 MJ kg^-1 and is at a radius of 8000 km. Surface B has a potential of -30 MJ kg^-1 and is at a radius of 10670 km. Estimate the gravitational field strength in the region between A and B.

  1. 1
    1. Use the potential gradient formula: g ≈ -ΔV_g / Δr.
  2. 2
    1. Calculate the change in potential, ΔV_g = V_final - V_initial = (-30 x 10^6) - (-40 x 10^6) = +10 x 10^6 J kg^-1.
  3. 3
    1. Calculate the change in radius, Δr = (10670 - 8000) km = 2670 km = 2.67 x 10^6 m.
  4. 4
    1. Substitute into the formula: g ≈ -(10 x 10^6) / (2.67 x 10^6).
  5. 5
    1. Calculate the value: g ≈ -3.75 N kg^-1. The negative sign indicates the field points radially inwards, so the magnitude is 3.75 N kg^-1.

Recap

  • An equipotential is a surface of constant gravitational potential.
  • No work is done moving a mass along an equipotential surface.
  • Gravitational field lines are always perpendicular to equipotential surfaces.
  • Field lines point in the direction of decreasing potential (more negative).
  • Field strength 'g' is equal to the negative of the potential gradient.
  • Equipotential lines are closer together where the field is stronger.

Quick check

  1. What is the angle between a gravitational field line and an equipotential surface at their point of intersection?1 mark
  2. If you move from a potential of -50 MJ/kg to -60 MJ/kg, are you moving towards or away from the source mass?1 mark

6. The Mechanics of Orbits

For a satellite to maintain a stable circular orbit around a planet, it must have a specific velocity. The force keeping it in this circular path is the gravitational force from the planet. This gravitational force provides the necessary centripetal force. By equating the formula for gravitational force with the formula for centripetal force (F = mv^2/r), we can derive expressions for the satellite's orbital speed and its period (the time for one full orbit). The derivation shows that the orbital speed 'v' depends only on the mass of the central body 'M' and the orbital radius 'r', not the mass of the satellite 'm'. A consequence of this is Kepler's Third Law, which states that the square of the orbital period (T^2) is directly proportional to the cube of the orbital radius (r^3).

G * M * m / r^2 = m * v^2 / r

v = sqrt(G * M / r)

T^2 = (4 * π^2 / (G * M)) * r^3

Key term

Geostationary Orbit: An orbit directly above the Earth's equator with a period of 24 hours, meaning the satellite remains at a fixed point above the Earth's surface.

Fun fact

The International Space Station travels at about 28,000 km/h. At that speed, the astronauts on board see a sunrise or a sunset every 45 minutes.

Worked example 14 marks

The International Space Station (ISS) orbits at an average altitude of 408 km. Given the Earth's mass is 5.97 x 10^24 kg and its radius is 6370 km, calculate the orbital speed of the ISS.

  1. 1
    1. First, find the total orbital radius 'r' from the center of the Earth.
  2. 2
    1. r = Earth's radius + altitude = (6370 km) + (408 km) = 6778 km = 6.778 x 10^6 m.
  3. 3
    1. Use the orbital speed formula: v = sqrt(G * M / r).
  4. 4
    1. Substitute the values: v = sqrt((6.67 x 10^-11 * 5.97 x 10^24) / (6.778 x 10^6)).
  5. 5
    1. Calculate the term inside the square root: (3.982 x 10^14) / (6.778 x 10^6) = 5.875 x 10^7.
  6. 6
    1. Take the square root: v = sqrt(5.875 x 10^7) = 7665 m s^-1 (or about 7.7 km s^-1).

Worked example 24 marks

Calculate the orbital radius for a geostationary satellite orbiting Earth. (A day is 86400 seconds).

  1. 1
    1. Start with the formula relating period and radius: T^2 = (4π^2 / GM) * r^3.
  2. 2
    1. Rearrange the formula to make r^3 the subject: r^3 = (G * M * T^2) / (4π^2).
  3. 3
    1. List the values: G = 6.67 x 10^-11, M = 5.97 x 10^24 kg, T = 86400 s.
  4. 4
    1. Substitute the values: r^3 = (6.67 x 10^-11 * 5.97 x 10^24 * (86400)^2) / (4π^2).
  5. 5
    1. Calculate the numerator: (3.982 x 10^14) * (7.465 x 10^9) = 2.973 x 10^24.
  6. 6
    1. Calculate the denominator: 4π^2 ≈ 39.48.
  7. 7
    1. Divide to find r^3: r^3 = (2.973 x 10^24) / 39.48 = 7.53 x 10^22 m^3.
  8. 8
    1. Take the cube root to find r: r = (7.53 x 10^22)^(1/3) = 4.22 x 10^7 m (or 42,200 km).

Recap

  • In a stable orbit, gravitational force provides the centripetal force.
  • Orbital speed decreases as the orbital radius increases.
  • The mass of the satellite does not affect its orbital speed or period.
  • The square of the orbital period is proportional to the cube of the orbital radius (T^2 ∝ r^3).
  • A geostationary satellite has a period of 24 hours and orbits above the equator.

Quick check

  1. If a satellite moves to a lower orbit (smaller r), does its orbital period increase or decrease?1 mark

7. Orbital Energy and Escape Velocity

An orbiting satellite possesses both kinetic energy (KE) due to its motion and gravitational potential energy (GPE) due to its position in the field. The total energy is the sum of these two. By using the formulas for orbital speed and GPE, we can show that the total energy of a satellite in a circular orbit is always negative and is given by E_total = -GMm/(2r). The negative sign is significant: it means the satellite is 'bound' to the planet and cannot escape without an input of energy. To escape the planet's gravitational pull completely (i.e., to reach infinity), an object must be given enough kinetic energy to overcome its negative GPE. The minimum speed required to do this from a point 'r' is called the escape velocity. This occurs when the object's initial kinetic energy is exactly equal to the magnitude of its initial GPE. This gives the formula v_esc = sqrt(2GM/r).

KE = (1/2) * G * M * m / r

E_total = KE + E_p = -G * M * m / (2r)

v_esc = sqrt(2 * G * M / r)

Key term

Escape Velocity: The minimum initial velocity an object must have to escape the gravitational field of a celestial body and never return, without any further propulsion.

Examiner insight

Questions about orbital energy changes are common. Remember that moving to a higher orbit requires work to be done, so total energy must increase (become less negative).

Common pitfall

Confusing the formula for orbital velocity, v = sqrt(GM/r), with the formula for escape velocity, v_esc = sqrt(2GM/r). Remember that you need more speed to escape than to orbit.

Worked example 13 marks

A 2000 kg satellite is in a stable orbit at a radius of 8.0 x 10^6 m around the Earth (M = 5.97 x 10^24 kg). Calculate its total energy.

  1. 1
    1. Use the formula for total energy in orbit: E_total = -GMm/(2r).
  2. 2
    1. Substitute the values: E_total = -(6.67 x 10^-11 * 5.97 x 10^24 * 2000) / (2 * 8.0 x 10^6).
  3. 3
    1. Calculate the numerator: -(7.964 x 10^17).
  4. 4
    1. Calculate the denominator: 1.6 x 10^7.
  5. 5
    1. Divide to find the total energy: E_total = -4.98 x 10^10 J.

Worked example 23 marks

Calculate the escape velocity from the surface of Mars. (Mass of Mars = 6.42 x 10^23 kg, Radius of Mars = 3.39 x 10^6 m).

  1. 1
    1. Use the formula for escape velocity: v_esc = sqrt(2GM/r).
  2. 2
    1. Substitute the values: v_esc = sqrt((2 * 6.67 x 10^-11 * 6.42 x 10^23) / (3.39 x 10^6)).
  3. 3
    1. Calculate the term inside the square root: (8.564 x 10^13) / (3.39 x 10^6) = 2.526 x 10^7.
  4. 4
    1. Take the square root: v_esc = sqrt(2.526 x 10^7) = 5026 m s^-1 (or 5.03 km s^-1).

Recap

  • The total energy of an orbiting body is the sum of its KE and GPE.
  • Total energy in orbit is always negative, signifying a bound system.
  • For a circular orbit, KE = -1/2 GPE and Total Energy = +1/2 GPE.
  • Escape velocity is the speed needed to make the total energy zero.
  • Escape velocity is sqrt(2) times the orbital velocity at the same radius.

Quick check

  1. What is the significance of a satellite having negative total energy?1 mark
  2. Does the escape velocity from a planet depend on the mass of the object escaping?1 mark

8. Orbital Decay and Atmospheric Drag

For satellites in Low Earth Orbit (LEO), the Earth's atmosphere, though extremely thin, still exerts a small frictional force known as atmospheric drag. This drag force acts in the opposite direction to the satellite's velocity, doing negative work on it. This causes the satellite's total energy to decrease (become more negative). Here comes the paradox: since the total energy is E_total = -GMm/(2r), a decrease in E_total means the orbital radius 'r' must also decrease. The satellite moves to a lower orbit. As the satellite drops to a lower orbit, its gravitational potential energy decreases significantly, while its kinetic energy increases (since v = sqrt(GM/r), a smaller 'r' means a larger 'v'). The decrease in GPE is twice the increase in KE, with the difference being dissipated as heat by the drag force. So, counter-intuitively, atmospheric drag causes a satellite to speed up as it spirals inwards towards the Earth, eventually burning up in the denser lower atmosphere.

Key term

Orbital Decay: The gradual decrease in the altitude of a satellite's orbit, typically caused by atmospheric drag.

Examiner insight

Examiners reward students who can clearly explain the counter-intuitive result that atmospheric drag causes a satellite to speed up as its altitude decreases. This is often called the 'satellite paradox'.

Common pitfall

Assuming that drag, a resistive force, must cause the satellite to slow down. While it removes total energy, this loss results in a lower orbit which has a higher kinetic energy and thus a higher speed.

Worked example 15 marks

A satellite is in a decaying orbit around the Earth due to atmospheric drag. Describe and explain the changes, if any, to the satellite's(a) total energy,(b) orbital radius, and(c) speed.

  1. 1
    1. (a) Total Energy: The drag force does negative work on the satellite, removing energy from the system. Therefore, the total energy of the satellite decreases (becomes more negative).
  2. 2
    1. (b) Orbital Radius: The total energy of an orbiting satellite is E = -GMm/(2r). As the total energy E decreases, the orbital radius 'r' must also decrease. The satellite spirals to a lower altitude.
  3. 3
    1. (c) Speed: The speed of a satellite is given by v = sqrt(GM/r). Since the orbital radius 'r' is decreasing, the orbital speed 'v' must increase. The satellite speeds up as it falls.

Recap

  • Atmospheric drag does negative work on a satellite in low orbit.
  • This causes the satellite's total energy to decrease.
  • As total energy decreases, the orbital radius also decreases.
  • As the orbital radius decreases, the satellite's speed increases.
  • This process is called orbital decay and leads to the satellite burning up.

Quick check

  1. What force is responsible for orbital decay?1 mark
  2. As a satellite's orbit decays, does its kinetic energy increase or decrease?1 mark

End-of-chapter exercise

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

  1. Two identical spherical asteroids, each with a mass of 5.0 x 10^12 kg, are at rest with their centers separated by 1000 m. Calculate the magnitude of the gravitational force that each asteroid exerts on the other.2 marks
  2. Jupiter has a mass of 1.90 x 10^27 kg and a radius of 6.99 x 10^7 m. Calculate the gravitational field strength at its surface.3 marks
  3. The mass of the Sun is 3.3 x 10^5 times the mass of the Earth. Their centers are separated by a distance D. Find the distance from the center of the Earth (in terms of D) where the net gravitational field is zero.5 marks
  4. A 1200 kg satellite orbits the Earth at an altitude where the gravitational potential is -2.5 x 10^7 J kg^-1. (a) Calculate the gravitational potential energy of the satellite. (b) How much energy must be supplied to move the satellite from this orbit to infinity?4 marks
  5. The gravitational potential near a planet is -8.0 x 10^8 J kg^-1 at a radius of 9.0 x 10^6 m and -7.2 x 10^8 J kg^-1 at a radius of 1.0 x 10^7 m. Estimate the gravitational field strength in this region.3 marks
  6. A spy satellite is in a circular orbit around the Earth at a radius of 7.0 x 10^6 m from the Earth's center. Calculate its orbital period in minutes. (Mass of Earth = 5.97 x 10^24 kg)4 marks
  7. Planet X has twice the mass and half the radius of Planet Y. Determine the ratio of the escape velocity from Planet X to the escape velocity from Planet Y (v_esc,X / v_esc,Y).4 marks
  8. A satellite of mass 500 kg is moved from an orbit of radius 2.0 x 10^7 m to an orbit of radius 3.0 x 10^7 m around a planet of mass 4.0 x 10^25 kg. Calculate the work done to move the satellite.4 marks
  9. Explain why the total energy of a comet in an elliptical orbit around the Sun is constant, but its kinetic and potential energies are not.3 marks
  10. A space station in a stable circular orbit fires its thrusters in the direction of motion, giving it a short boost of speed. Describe and explain the new orbit of the space station.4 marks

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