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Questions on Newton's law of gravitation
Newton's law of gravitation makes every pair of masses attract with a force Gm₁m₂/r², so doubling their separation quarters the pull. There is no repulsive case, because mass has one sign. Dividing out the test mass gives the gravitational field strength g = GM/r², with r measured from the centre of the body.
19 original questions · 53 marks · the Newton's law of gravitation notes · Gravitational fields
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State Newton's law of gravitation for two point masses.
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The force is attractive and proportional to the product of the masses (1); and inversely proportional to the square of their separation, F = GMm/r2 (1).Two point masses, each of mass 5.0 kg, are 2.0 m apart. Calculate the gravitational force between them.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
F = GMm/r2 = (6.67 × 10−11 × 5.0 × 5.0)/2.02 (1)
F = 4.17 × 10−10 N (1)Define gravitational field strength and write down the equation relating it to the force on a mass.
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Gravitational force per unit mass on a small test mass at the point (1); g = F/m (1).The Earth pulls a student of mass 60 kg with a gravitational force of 590 N. State the magnitude of the gravitational force the student exerts on the Earth, and explain why the Earth's resulting motion is negligible.
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590 N: by Newton's third law the two forces are equal in magnitude (and opposite in direction) (1). The Earth's mass is enormous, so its acceleration a = F/M is far too small to notice (1).The gravitational field strength at the Earth's surface is 9.8 N kg−1. Without using values of G or the Earth's mass, determine the field strength at a point 3 Earth radii from the Earth's centre.
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g follows an inverse-square law, so trebling the distance from the centre divides g by 32 = 9 (1)
g = 9.8/9 = 1.1 N kg−1 (1)State what is meant by a point mass, and explain why Newton's law of gravitation can be applied to spherical planets.
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A mass whose size is negligible compared with the separation involved, so that it can be treated as being at a single point (1). A (uniform) sphere behaves, from outside, as if all its mass were concentrated at its centre, so the law applies with r measured from the centre (1).A planet of mass 6.0 × 1024 kg and its moon of mass 7.3 × 1022 kg are separated by a distance of 3.8 × 108 m. Calculate the gravitational force between them.
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F = GMm/r2 = (6.67 × 10−11 × 6.0 × 1024 × 7.3 × 1022)/(3.8 × 108)2 (1)
F = 2.02 × 1020 N (1)A planet has a mass of 6.0 × 1024 kg and a radius of 6.4 × 106 m. Calculate the gravitational field strength at its surface.
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g = GM/r2 = (6.67 × 10−11 × 6.0 × 1024)/(6.4 × 106)2 (1)
g = 9.77 N kg−1 (1)For the same planet, calculate the gravitational field strength at a distance of 1.28 × 107 m from its centre. Comment on how your value compares with the surface value.
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g = GM/r2 = (6.67 × 10−11 × 6.0 × 1024)/(1.28 × 107)2 (1)
g = 2.44 N kg−1 (1)
This is one quarter of the surface value: the distance has doubled, so g falls by a factor of 4 (inverse-square law) (1).A satellite of mass 500 kg orbits a planet of mass 6.0 × 1024 kg at a distance of 7.0 × 106 m from its centre. Calculate the gravitational field strength at this distance and hence the gravitational force on the satellite.
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g = GM/r2 = (6.67 × 10−11 × 6.0 × 1024)/(7.0 × 106)2 = 8.17 N kg−1 (2)
F = mg = 500 × 8.17 = 4085 N ≈ 4.1 × 103 N (1)Two snooker balls, each of mass 74 g, rest with their centres 6.0 cm apart. Calculate the gravitational force each ball exerts on the other, and suggest why this force has no noticeable effect.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
Conversions: m = 0.074 kg, r = 0.060 m (1)
F = Gm2/r2 = (6.67 × 10−11 × 0.0742)/0.0602 = 1.0 × 10−10 N (1)
This is negligible compared with the other forces on a ball, such as its weight (about 0.7 N) or friction from the table (1).A planet of mass 6.4 × 1023 kg orbits a star of mass 2.0 × 1030 kg at a distance of 2.3 × 1011 m. Calculate the gravitational force of the star on the planet, and the acceleration of the planet.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
F = GMm/r2 = (6.67 × 10−11 × 2.0 × 1030 × 6.4 × 1023)/(2.3 × 1011)2 (1)
F = 1.61 × 1021 N (1)
a = F/m = 1.61 × 1021/(6.4 × 1023) (1)
a = 2.5 × 10−3 m s−2 (1)The gravitational field strength at a point above a planet of mass 6.4 × 1023 kg is 1.0 N kg−1. Determine the distance of this point from the planet's centre.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
r = √(GM/g) (1)
r = √((6.67 × 10−11 × 6.4 × 1023)/1.0) (1)
r = 6.5 × 106 m (1)Mars has a mass of 6.4 × 1023 kg and a radius of 3.4 × 106 m. Show that the gravitational field strength at the surface of Mars is approximately 3.7 N kg−1.
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g = GM/r2 (1)
g = (6.67 × 10−11 × 6.4 × 1023)/(3.4 × 106)2 (1)
g = 3.69 N kg−1 ≈ 3.7 N kg−1 (1)The gravitational field strength at the surface of a planet of radius 6.4 × 106 m is 9.81 N kg−1. Determine the mass of the planet.
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M = gr2/G (1)
M = (9.81 × (6.4 × 106)2)/(6.67 × 10−11) (1)
M = 6.02 × 1024 kg (1)Explain why the gravitational field strength of a planet decreases with distance from its centre, and state the mathematical form of this decrease.
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The field lines spread out over a larger area as distance increases, so the field becomes weaker (1). From g = GM/r2, g follows an inverse-square law with distance from the centre (1); for example, doubling r reduces g to one quarter of its value (1).A magazine article describes two newly discovered planets orbiting another star:
planet K: mass 3.0 × 1025 kg, radius 1.6 × 107 m
planet L: mass 8.9 × 1024 kg, radius 7.7 × 106 m
The article claims that both planets have a surface gravitational field strength greater than the Earth's value of 9.8 N kg−1. Deduce whether the claim is correct.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
Use of g = GM/r2 for each planet (1)
gK = (6.67 × 10−11 × 3.0 × 1025)/(1.6 × 107)2 = 7.8 N kg−1 (1)
gL = (6.67 × 10−11 × 8.9 × 1024)/(7.7 × 106)2 = 10.0 N kg−1 (1)
The claim is not correct: only L exceeds 9.8 N kg−1; K's larger mass is outweighed by its larger radius (1).Estimate the gravitational force of attraction between two people sitting 1.0 m apart, and estimate how many orders of magnitude smaller this is than your weight.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
Sensible estimate of the masses, for example 70 kg each (1)
F = Gm1m2/r2 = (6.67 × 10−11 × 70 × 70)/1.02 ≈ 3.3 × 10−7 N (1)
This is of order 10−7 N against a weight of about 700 N: some nine orders of magnitude smaller (1).A crewed space laboratory orbits 4.2 × 105 m above the surface of a planet of mass 6.0 × 1024 kg and radius 6.4 × 106 m. Show that the gravitational field strength at this altitude is about 8.6 N kg−1, and explain why the crew float inside the laboratory even though g there is far from zero.
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r = 6.4 × 106 + 4.2 × 105 = 6.82 × 106 m, measured from the planet's centre (1)
g = GM/r2 = (6.67 × 10−11 × 6.0 × 1024)/(6.82 × 106)2 (1)
g = 8.60 N kg−1 ≈ 8.6 N kg−1 (1)
The laboratory and the crew are both in free fall, accelerating towards the planet at the same rate (1); there is therefore no contact (support) force between the crew and the laboratory, so the crew appear weightless (1).
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