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A gravitational field is a region of space in which a mass experiences a force due to the presence of another mass. Every object with mass creates a gravitational field around it. Understanding gravitational fields is essential for A-Level Physics (AQA specification 3.7.1) and forms the foundation for orbital mechanics, satellite motion, and astrophysics.
Gravitational field lines are used to represent the direction and relative strength of a gravitational field. They are drawn as arrows that point in the direction a small test mass would accelerate if placed at that point.
Described diagram — Field lines around a spherical mass: Lines are drawn radially inward towards the centre of the mass, uniformly distributed around its surface. The lines are closer together near the surface (indicating a stronger field) and spread further apart at greater distances (indicating a weaker field). The arrows on every line point towards the centre of the mass, because gravity is always attractive.
Key properties of gravitational field lines:
Near the surface of a large body such as the Earth, over small distances, the gravitational field is approximately uniform. This means the field strength has the same magnitude and direction at every point in the region.
Described diagram — Uniform gravitational field: Equally spaced, parallel, vertical lines pointing downward. The uniform spacing indicates that the field strength is the same everywhere in the region.
In a uniform field:
Near the Earth's surface, g ≈ 9.81 N kg⁻¹ (equivalently, 9.81 m s⁻²). This approximation is valid for heights much smaller than the Earth's radius (6.37 × 10⁶ m).
Exam Tip: The units of gravitational field strength are N kg⁻¹, which are dimensionally equivalent to m s⁻². Both are acceptable, but N kg⁻¹ is preferred when discussing field strength, while m s⁻² is preferred when discussing acceleration. The conceptual distinction matters: g as field strength is a property of the field at a point; g as acceleration is the response of a mass placed in that field.
Key Definition: The gravitational field strength (g) at a point is the force per unit mass acting on a small test mass placed at that point.
g = F/m
where g is the gravitational field strength (N kg⁻¹), F is the gravitational force (N), and m is the mass of the test mass (kg).
The definition uses a "small test mass" to ensure that the test mass does not significantly disturb the field it is measuring. If the test mass were very large, its own gravitational field would alter the field being measured.
Gravitational field strength is a vector quantity — it has both magnitude and direction. The direction is always towards the mass creating the field.
Key Definition: Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
F = GMm/r²
where:
Key features:
Combining g = F/m with F = GMm/r²:
g = F/m = (GMm/r²)/m = GM/r²
g = GM/r²
This gives the gravitational field strength at a distance r from the centre of a mass M. For a uniform sphere, this formula applies outside the sphere (r ≥ R, where R is the radius of the sphere), and the mass behaves as if concentrated at the centre.
Question: The Earth has a mass of 5.97 × 10²⁴ kg and a radius of 6.37 × 10⁶ m. Calculate the gravitational field strength at the Earth's surface.
Solution:
Using g = GM/r²:
g = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.37 × 10⁶)²
g = (3.98 × 10¹⁴) / (4.06 × 10¹³)
g = 9.81 N kg⁻¹
This confirms the familiar value of g at the Earth's surface.
Question: Calculate the gravitational force between the Earth (mass 5.97 × 10²⁴ kg) and the Moon (mass 7.35 × 10²² kg). The mean Earth–Moon distance is 3.84 × 10⁸ m.
Solution:
Using F = GMm/r²:
F = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²²) / (3.84 × 10⁸)²
Numerator: 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ = 3.98 × 10¹⁴ 3.98 × 10¹⁴ × 7.35 × 10²² = 2.93 × 10³⁷
Denominator: (3.84 × 10⁸)² = 1.47 × 10¹⁷
F = 2.93 × 10³⁷ / 1.47 × 10¹⁷ = 1.99 × 10²⁰ N
This enormous force (approximately 2 × 10²⁰ N) is what keeps the Moon in orbit around the Earth.
Question: A satellite orbits at an altitude of 400 km above the Earth's surface. Calculate the gravitational field strength at the satellite's orbital altitude. (Earth mass = 5.97 × 10²⁴ kg, Earth radius = 6.37 × 10⁶ m.)
Solution:
The distance from the centre of the Earth is: r = 6.37 × 10⁶ + 400 × 10³ = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m
g = GM/r² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.77 × 10⁶)²
g = 3.98 × 10¹⁴ / 4.58 × 10¹³ = 8.69 N kg⁻¹
Note that g has only decreased from 9.81 to 8.69 N kg⁻¹ — a reduction of about 11%. Astronauts aboard the International Space Station (which orbits at approximately this altitude) experience "weightlessness" not because gravity is zero, but because both the station and the astronauts are in free fall together.
Common Misconception: Students often believe that astronauts float in the ISS because there is no gravity in space. In fact, gravity at the ISS orbit is about 89% as strong as at the surface. The apparent weightlessness occurs because the ISS and everything inside it are in continuous free fall — they are all accelerating towards the Earth at the same rate.
When more than one mass is present, the total gravitational field at any point is the vector sum of the individual fields due to each mass. This is the principle of superposition.
Question: The Earth has mass M_E = 5.97 × 10²⁴ kg and the Moon has mass M_M = 7.35 × 10²² kg. The distance between their centres is d = 3.84 × 10⁸ m. Find the distance from the Earth's centre to the point where the gravitational field strength is zero (the neutral point).
Solution:
At the neutral point, the field due to the Earth equals the field due to the Moon (in magnitude but opposite in direction).
GM_E/x² = GM_M/(d − x)²
Cancelling G:
M_E/x² = M_M/(d − x)²
Taking square roots:
√M_E / x = √M_M / (d − x)
(d − x)/x = √(M_M/M_E) = √(7.35 × 10²² / 5.97 × 10²⁴) = √(0.01231) = 0.1109
d − x = 0.1109x
d = 1.1109x
x = d/1.1109 = 3.84 × 10⁸ / 1.1109 = 3.46 × 10⁸ m from the Earth's centre.
This is about 90% of the way to the Moon, confirming that the Earth's much larger mass dominates most of the space between them.
| Concept | Formula | Units |
|---|---|---|
| Gravitational field strength (definition) | g = F/m | N kg⁻¹ |
| Newton's law of gravitation | F = GMm/r² | N |
| Field strength for a point/spherical mass | g = GM/r² | N kg⁻¹ |
| Gravitational constant | G = 6.67 × 10⁻¹¹ | N m² kg⁻² |
Exam Tip: In multi-mark calculations, always show your substitution clearly and include units at every stage. State the formula, substitute, then evaluate. Even if you make an arithmetic error, you will gain method marks for correct setup and substitution.