OxfordAQA International A Level Physics
OxfordAQA International A Level Physics 3.7 Gravitational Fields and Satellites Exam Questions
Free OxfordAQA International A Level Physics revision for 3.7 Gravitational Fields and Satellites: what you need to know, common mistakes and exam-style questions with answers. Covers newton's law of gravitation, gravitational field strength, gravitational potential, equipotentials, orbital period and speed, orbital energy and geosynchronous satellites.
Practise 3.7 Gravitational Fields and Satellites with past papers
Once you have learnt this topic, find the 3.7 Gravitational Fields and Satellites questions in real OxfordAQA International A Level Physics past papers and mark them with the mark scheme. Doing lots of real questions is the quickest way to learn how examiners ask about each idea.
About OxfordAQA International A Level Physics 3.7 Gravitational Fields and Satellites
Gravitational fields and satellites is part of Unit 3 (PH03), Fields and their consequences, in the OxfordAQA International A2 year. You use Newton's law of gravitation to describe the force between point masses, define gravitational field strength and gravitational potential, and apply these to the orbits of planets and satellites. It is closely parallel to electric fields, and examiners often ask you to compare the two. Typical questions include deriving the relationship between orbital period and radius, calculating the radius of a geosynchronous orbit, sketching g and V against distance, and finding the energy needed to move a mass between two points in a field.
What you need to know
- 3.7.1 Newton's law F = Gm1m2 / r^2, an attractive force between point masses, where G = 6.67 x 10^-11 N m^2 kg-2 and r is the distance between centres.
- 3.7.2 Field strength g = F / m is the force per unit mass. Near a spherical mass, g = GM / r^2 (radial field). Close to the Earth's surface the field is approximately uniform.
- 3.7.3 Gravitational potential V is the work done per unit mass in bringing a small mass from infinity to that point; V = -GM / r, zero at infinity and always negative. Work done moving a mass m = m delta V. Equipotential surfaces are perpendicular to field lines, and g = -delta V / delta r, so g is the negative gradient of a V-r graph.
- 3.7.4 Orbits Gravity provides the centripetal force: GMm / r^2 = mv^2 / r, giving v^2 = GM / r and T^2 = 4 pi^2 r^3 / GM, so T^2 is proportional to r^3.
- Orbital energy Total energy of an orbiting satellite is kinetic plus potential energy. Moving to a higher orbit requires energy input; the satellite then moves more slowly.
- Geosynchronous orbits A geosynchronous satellite has a period of one day, orbiting at about 3.6 x 10^7 m above the surface. A geostationary orbit is above the equator, moving in the same direction as the Earth rotates, so it stays above one point, useful for communications.
Common mistakes in 3.7 Gravitational Fields and Satellites
- Using the distance from the surface instead of from the centre of the planet in g = GM / r^2 or V = -GM / r. Always add the planet's radius to the height.
- Giving gravitational potential as positive. It is negative everywhere and rises to zero at infinity.
- Saying there is no gravity on an orbiting astronaut. Gravity is still acting; the astronaut is in free fall.
- Thinking a satellite in a higher orbit moves faster. Orbital speed decreases as orbital radius increases.
- Forgetting to convert a period in hours or days into seconds before using T^2 = 4 pi^2 r^3 / GM.
Try three 3.7 Gravitational Fields and Satellites exam-style questions
1. Show that the orbital period T of a satellite is related to its orbital radius r by T^2 = 4 pi^2 r^3 / GM. (3 marks)
Show the mark scheme answer
Gravitational force provides the centripetal force: GMm / r^2 = mv^2 / r; v = 2 pi r / T, so GM / r = 4 pi^2 r^2 / T^2; rearranging gives T^2 = 4 pi^2 r^3 / GM.
2. Calculate the orbital radius of a satellite with a period of 24 hours. Mass of Earth = 6.0 x 10^24 kg, G = 6.67 x 10^-11 N m^2 kg-2. (3 marks)
Show the mark scheme answer
T = 24 x 3600 = 86400 s; r^3 = GMT^2 / 4 pi^2 = 6.67 x 10^-11 x 6.0 x 10^24 x 86400^2 / 39.5 = 7.6 x 10^22 m^3; r = 4.2 x 10^7 m.
3. Explain why the gravitational potential at a point near the Earth is negative. (2 marks)
Show the mark scheme answer
Potential is defined as zero at infinity; gravitational force is attractive so work must be done on a mass to move it from the point to infinity (energy is released bringing it in), so the potential at the point is less than zero.
3.7 Gravitational Fields and Satellites FAQs
What does OxfordAQA International A Level Physics 3.7 Gravitational Fields and Satellites cover?
It covers Newton's law of gravitation, gravitational field strength, gravitational potential and equipotentials, orbital speed and period, orbital energy and geosynchronous orbits.
Which unit is 3.7 Gravitational Fields and Satellites examined in?
It is examined in Unit 3 (PH03), Fields and their consequences.
How is gravitational potential linked to field strength?
Field strength is the negative of the potential gradient, g = -delta V / delta r. On a graph of V against r, the gradient at any point gives the magnitude of g there.