- Use the law of gravitation and the height dependence of g in ratio problems
- Tell the force of gravity from weight, and mass from weight
- Calculate weight in an accelerating lift and on a bridge
- Explain weightlessness and g-load
Stand on bathroom scales in the lift of a tall building. When the lift starts going up the scales show more, and when it starts going down they show less — yet your mass has not changed. An astronaut on a space station is even “weightless”, although the Earth still pulls on them. In this lesson you will learn to tell the force of gravity from weight and to calculate when weight grows or shrinks.
Universal gravitation and the force of gravity
Newton found that every body in the Universe attracts every other body. The gravitational force is proportional to the product of the masses and inversely proportional to the square of the distance between their centres. The force that makes an apple fall also keeps the Moon in orbit. Planets and satellites are covered in the lesson «Circular motion and gravitation».
- Fgravitational force, in N
- m₁, m₂masses of the bodies, in kg
- rdistance between the centres, in m
- Ggravitational constant, 6.67 · 10⁻¹¹ N · m² / kg²: the pull between two 1 kg bodies 1 m apart
The force with which the Earth pulls bodies towards itself is the force of gravity. It acts on the body itself, points vertically down towards the Earth’s centre and is proportional to the mass. At the Earth’s surface g = G · M / R² ≈ 9.8 N/kg (10 N/kg in problems). Further from the Earth, g gets smaller.
- mmass of the body, in kg
- g₀free-fall acceleration at the Earth’s surface
- ghfree-fall acceleration at height h
- Rradius of the Earth (≈ 6400 km)
The force of gravity and how it depends on height.
The Earth creates a gravitational field around itself. The field strength at a point is the force on a 1 kg body placed there: g = F / m (N/kg). So g has two meanings: the field strength (N/kg) and the free-fall acceleration (m/s²) — the units are the same, because 1 N/kg = 1 m/s².
Murad has a mass of 50 kg. Find the force of gravity on him on the Earth (g = 9.8 N/kg) and on the Moon (g = 1.6 N/kg). What is Murad’s mass on the Moon?
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Mass is a property of the body itself and does not change: it is still 50 kg on the Moon.
1) The distance between two bodies is tripled. How does the gravitational force change?
2) What is g at a height R above the Earth’s surface (R is the Earth’s radius; g₀ = 10 m/s²)?
3) At what height is the free-fall acceleration 9 times smaller than at the surface?
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2) g = g₀ · R² / (2R)² = g₀ / 4 = 2.5 m/s².
3) (R + h)² / R² = 9 → R + h = 3R → h = 2R (≈ 12,800 km).
Weight and mass
The force (W) with which a body, pulled by the Earth, presses on its support or pulls on its hanger. Weight acts on the support, while the force of gravity acts on the body itself. By Newton’s third law the weight equals the normal force: W = N. At rest or in uniform motion W = mg. (Many English textbooks simply call mg the weight; here we use the stricter meaning that DİM uses.)
| Property | Mass | Weight |
|---|---|---|
| What is it? | a measure of inertia | a force on the support |
| Unit | kg | N |
| Instrument | balance | spring balance |
| On the Moon or in a lift | stays the same | changes |
Weight in an accelerating lift
Two forces act on a body in a lift: mg (down) and the normal force N (up). If the acceleration points up, Newton’s second law gives N − mg = ma, so N = m(g + a). If it points down, N = m(g − a). The weight is W = N. The ratio of the weight to mg is called the g-load (overload) n.
- amagnitude of the lift’s acceleration, in m/s²
- ng-load: n > 1 — weight increased, n = 0 — weightlessness
| Motion of the lift | Acceleration | Weight |
|---|---|---|
| speeds up going up | up | W > mg |
| moves uniformly or stands still | a = 0 | W = mg |
| brakes while going up | down | W < mg |
| speeds up going down | down | W < mg |
| brakes while going down | up | W > mg |
| falls freely | a = g | W = 0 |
Aysel, whose mass is 50 kg, stands on scales in a lift. What do the scales show (in newtons) when the lift a) starts going up with an acceleration of 2 m/s², b) starts going down with an acceleration of 2 m/s², c) moves at constant speed? What is the g-load in case a)?
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n = 600 / 500 = 1.2 — Aysel feels 1.2 times “heavier”.
A lift moving down brakes with a deceleration of 3 m/s² as it approaches a floor. Find the weight of a 70 kg person in the lift (g = 10 m/s²).
A) 490 N B) 700 N C) 910 N D) 210 N E) 1400 N
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W = m(g + a) = 70 · 13 = 910 N (C).
A (490 N) is the mistake of looking at the velocity and writing W = m(g − a).
A car on a convex or concave bridge
If a bridge is an arc of a circle, the car moves along a circle and its centripetal acceleration a = v² / R points to the centre (see «Circular motion and gravitation»). At the top of a convex bridge the centre is below — the acceleration points down and the weight drops. At the bottom of a concave bridge the centre is above — the weight grows.
- vspeed of the car, in m/s
- Rradius of curvature of the bridge, in m
When v = √(g · R), W = 0 at the top of a convex bridge — weightlessness.
A 2 t car drives at 15 m/s over a bridge with a radius of curvature of 45 m. With what force does it press on the bridge a) at the top of a convex bridge, b) at the bottom of a concave bridge? c) At what speed would the car be weightless at the top of the convex bridge?
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a) W = 2000 · (10 − 5) = 10 kN.
b) W = 2000 · (10 + 5) = 30 kN.
c) v = √(g · R) = √450 ≈ 21 m/s.
Weightlessness and g-load
When a body and its support move under gravity alone (a = g, down), the body does not press on the support: W = m(g − g) = 0. This is weightlessness. A jumping athlete in the air, a freely falling lift and a space station in orbit are all weightless: the station is constantly “falling” around the Earth. The force of gravity still acts! During a rocket launch or a tight aircraft turn, on the contrary, a g-load (n > 1) appears.
An aircraft flies a vertical loop of radius 500 m and passes its lowest point at 100 m/s. Find the force with which the 80 kg pilot presses on the seat and the g-load.
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W = m(g + a) = 80 · 30 = 2400 N; n = 2400 / 800 = 3 — the pilot feels three times “heavier”.
Which statements are true? I. In weightlessness the force of gravity on a body is zero. II. Weight acts on the support or the hanger. III. In a lift accelerating upwards a body’s weight is greater than mg.
A) only I B) only II C) II and III D) I and III E) I, II, III
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Answer: C.
Key points
- F = G · m₁ · m₂ / r²; the force of gravity is F = mg, and g falls with height: gh = g₀R² / (R + h)².
- Gravity acts on the body, weight acts on the support; at rest the weight equals mg.
- In a lift the weight is m(g ± a): “+” for acceleration up, “−” for down; the direction of the velocity does not matter.
- On a convex bridge the weight is m(g − v²/R); on a concave one, m(g + v²/R).
- In weightlessness (a = g) the weight is zero, but gravity still acts.
Check yourself
12 questions. Every correct answer earns XP.