Exam practice

g = GM/R², so g depends on the distance from the Earth’s center. The Earth is not a perfect sphere: it is flattened at the poles and bulges at the equator, so places at the poles are closer to the center than places at the equator. Mountain tops are farther from the center than the plains. So g differs from place to place: about 9.83 m/s² at the poles, 9.78 m/s² at the equator, and a little less on high mountains. (The Earth’s spin also reduces g slightly at the equator.)

A falling body keeps speeding up: from v² = u² + 2gh with u = 0, its speed on landing is v = √(2gh), which is larger the greater the height. A larger speed has to be brought to zero in the same short moment of hitting the ground, which needs a much larger force on the body (the change of momentum is larger), so the injury is greater. From 1 m you land at 4.4 m/s; from 5 m at 9.9 m/s.

g = GM/R² depends on the radius as well as the mass, and on the square of the radius. Jupiter’s radius is about 11 times the Earth’s, so R² is 121 times as large. So its g is 319/121 ≈ 2.6 times the Earth’s: the effect of its huge mass is largely canceled by its huge size.

The Earth is flattened at the poles, so the polar regions are closer to the Earth’s center than the equator. Since g ∝ 1/R², g is larger at the poles (9.83 m/s²) than at the equator (9.78 m/s²). With a larger acceleration, the object at the pole speeds up faster and reaches the ground first.

Both have the same mass and weight, but the flat sheet has a much larger surface facing the air, so it meets much more air resistance, which slows it down. The ball is compact, so air resistance on it is small, and it falls almost freely. So the ball reaches the ground first.

In a vacuum there is no air resistance, so both fall freely with acceleration g. Since g = GM/R² does not depend on the mass of the falling object, the marble and the feather speed up at the same rate and land together.

As you climb, you move farther from the Earth’s center, so g = GM/(R+h)² decreases (from about 9.81 m/s² at sea level to about 9.78 m/s² at the summit). Weight W = mg, so with the same mass, the weight decreases a little. (Only by about 0.3%: the load won’t feel lighter, because the climber gets tired and the air is thin!)

To lift a stone, you must pull up with a force at least equal to its weight. Weight W = mg, and at the same place g is the same, so weight is proportional to mass. A big stone has more mass, so more weight, and needs a bigger force to lift.

Mass is the quantity of matter in the object, which doesn’t change when the object is moved. Weight is the force of gravity on it, W = mg, and g changes with place (the poles, the equator, mountain tops, the Moon). So the weight changes while the mass stays the same.

As the seat moves down over the top of the wheel, it accelerates downwards. You fall with it, nearly freely, so the seat presses up on you much less than usual: you are partly weightless. Your organs, normally held up by your body, also press less on each other, and that is the strange, floating feeling in your stomach.

Gravity (the force of gravity) is the force with which a planet, moon or star pulls an object towards its center. The Earth’s gravity is what gives objects their weight and makes them fall.

Every object in the universe attracts every other object 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: $F = \dfrac{G m_1 m_2}{d^2}$.

• It is always attractive, never repulsive.
• It acts between all objects with mass, anywhere in the universe.
• It is mutual: the two objects pull each other with equal and opposite forces.
• It acts along the line joining the centers of the two objects.
• It is very weak for small objects, but very large for planets and stars.
• It acts across any distance, even through empty space, and doesn’t depend on the material between the objects.

G is the gravitational force between two objects of mass 1 kg each whose centers are 1 m apart. Its value is 6.67 × 10⁻¹¹ N m²/kg².

When m₁ = m₂ = 1 kg and d = 1 m: then $F = \dfrac{G \times 1 \times 1}{1^2} = G$.
(More generally, whenever m₁m₂/d² = 1.)

1. It keeps the planets moving round the Sun, and the Moon round the Earth.
2. The Moon’s gravitational pull causes the tides in the oceans.

At first: $F_1 = \dfrac{G m_1 m_2}{d^2}$.
New masses 2m₁ and 2m₂; new distance d/4:
$$F_2 = \frac{G (2m_1)(2m_2)}{(d/4)^2} = \frac{4 G m_1 m_2}{d^2/16} = 64 \times \frac{G m_1 m_2}{d^2} = 64 F_1$$
The force becomes 64 times as large.

The gravitational force (gravitation) is the force of attraction between any two objects in the universe because of their masses. For example, the force between the Sun and the Earth, or between two stones.

The acceleration produced in a freely falling object by the force of gravity is called the acceleration due to gravity (g). On the Earth’s surface it is about 9.8 m/s².

Free fall is the motion of an object falling under the force of gravity alone, with no other force (such as air resistance) acting on it.
Examples:

• i.the hammer and the feather dropped on the Moon, where there is no air;
• ii.the coin and the feather in a tube with the air pumped out.

Very nearly free fall: a stone dropped from a short height; the Space Station, and the astronauts in it, orbiting the Earth.

When the only force acting on it is gravity: there is no air resistance or upthrust (as in a vacuum, or where air resistance is negligible), and nothing holds it up or pushes it. Its acceleration is then g.

1. In air, the coin falls faster than the feather, because air resistance slows the feather much more.
2. In a vacuum, the feather and the coin fall together.
3. So, without air resistance, all objects fall with the same acceleration: g does not depend on the mass of the falling object.

Weightlessness is the state of an object in free fall, in which it presses on nothing and a spring balance falling with it reads zero, so it seems to have no weight. Gravity still acts on it, but everything around it falls with the same acceleration. Astronauts in an orbiting spacecraft are weightless.

1. Every object has weight.
2. The Earth holds its atmosphere.
3. Objects fall to the ground when dropped, and things thrown up come back down.
4. Rivers and streams flow downhill.

(Also: falling objects accelerate; rain falls; we stay on the ground.)

Let an object of mass m be on the surface of the Earth, of mass M and radius R. By Newton’s law of gravitation, the force on it is $F = \dfrac{GMm}{R^2}$ …(i).
This force gives it an acceleration g, so by Newton’s second law, $F = mg$ …(ii).
From (i) and (ii): $mg = \dfrac{GMm}{R^2}$, so $g = \dfrac{GM}{R^2}$.
G is a constant, and the mass M of the Earth is fixed, so $g \propto \dfrac{1}{R^2}$.

From g = GM/(R + h)²:

1. The mass of the planet or moon (more mass, larger g).
2. The radius of the planet (larger radius, smaller g), so on Earth the shape: g is larger at the poles than at the equator.
3. The height above the surface (higher, smaller g).

Also, slightly, the Earth’s spin (it reduces g at the equator) and the density of the rocks below. It does not depend on the mass of the falling object.

It means that a freely falling object near the Earth’s surface gains a speed of 9.8 m/s every second (and an object thrown straight up loses 9.8 m/s every second). It also means the Earth pulls each kilogram of mass with a force of 9.8 N.

Only the radius changes: R becomes R/3.7.

Now: $g = \dfrac{GM}{R^2}$. Squeezed: the mass M is the same, and the radius is R/3.7:
$$g' = \frac{GM}{(R/3.7)^2} = 3.7^2 \times \frac{GM}{R^2} = 13.7\,g$$
g would become about 13.7 times as large: 13.7 × 9.8 ≈ 134 m/s². (The Moon’s mass, 1/81 of the Earth’s, isn’t needed: the question keeps the Earth’s mass. For comparison, the Moon’s own g is $\tfrac{1}{81} \times 3.7^2 = 0.17$, about a sixth of the Earth’s.)

Also 2 m/s². The acceleration due to gravity at a point is g = GM/d², which depends only on the mass M of the body pulling and the distance d from its center, not on the mass of the falling object. The 10 kg object is pulled 10 times harder, but it is also 10 times harder to speed up, so its acceleration is the same.

The mass will be the same in both places, as the amount of matter doesn’t change.
The weight will be slightly less in the mountains than in the Terai, because the mountains are farther from the Earth’s center, so g is smaller there, and W = mg.

No, it would not work. A beam balance compares the object with standard masses on the other pan. Moving to the Terai, g increases for the oranges and for the standard masses by exactly the same factor, so the beam balances at the same reading. A beam balance measures mass, which doesn’t change, so 10 kg bought is 10 kg sold, and there is no profit.
(With a digital scale it would be different. A digital or spring scale measures weight and shows it as kilograms. In the mountains, where g is a little smaller, the oranges press less and read slightly lighter; in the Terai the same oranges read slightly heavier, so the trick would “work”, but only by about 0.1% for a 3000 m difference in height: about 10 g on 10 kg. And a scale set up for one place should be adjusted for another.)

When the parachute opens, its large canopy meets a large air resistance, which grows with speed. Soon the air resistance becomes equal to the parachutist’s weight. The net force is then zero, so the acceleration is zero, and the parachutist falls at a small, steady speed (the terminal velocity, about 5 m/s), which allows a safe landing.
On the Moon, it isn’t possible: there is no atmosphere, so there is no air resistance. The parachute can’t slow the fall; the astronaut would be in free fall, keep speeding up, and hit the surface fast. (Spacecraft land on the Moon using rockets.)

By Newton’s second law, a = F/m: for a fixed force, a bigger mass gets a smaller acceleration. But in a fall, the force is not fixed: it is the weight, F = mg, which is itself proportional to the mass. A heavier object is pulled harder in exact proportion to its mass, so a = mg/m = g: the mass cancels, and every object falls with the same acceleration.

$$F = \frac{G m_1 m_2}{d^2} = \frac{6.67 \times 10^{-11} \times 20 \times 40}{5^2} = \frac{6.67 \times 10^{-11} \times 800}{25} = 2.134 \times 10^{-9}\ \text{N}$$
F = 2.13 × 10⁻⁹ N

d is between the centers: add both radii to the 5 m gap.

Given: m₁ = 5 kg, radius 10 cm = 0.10 m; m₂ = 2.5 kg, radius 5 cm = 0.05 m; gap between the surfaces 5 m.
Distance between the centers: d = 0.10 + 5 + 0.05 = 5.15 m.
$$F = \frac{6.67 \times 10^{-11} \times 5 \times 2.5}{5.15^2} = \frac{83.38 \times 10^{-11}}{26.52} = 3.14 \times 10^{-11}\ \text{N}$$
F = 3.14 × 10⁻¹¹ N. (Using 5 m by mistake gives 3.34 × 10⁻¹¹ N.)

d = 7.8 × 10⁸ km = 7.8 × 10¹¹ m.
$$F = \frac{6.67 \times 10^{-11} \times 2 \times 10^{30} \times 1.9 \times 10^{27}}{(7.8 \times 10^{11})^2} = \frac{25.35 \times 10^{46}}{60.84 \times 10^{22}} = 4.17 \times 10^{23}\ \text{N}$$
F = 4.17 × 10²³ N.

d = 3.84 × 10⁸ m. From $F = \dfrac{G m_1 m_2}{d^2}$:
$$m_2 = \frac{F d^2}{G m_1} = \frac{1.98 \times 10^{20} \times (3.84 \times 10^{8})^2}{6.67 \times 10^{-11} \times 5.972 \times 10^{24}} = \frac{1.98 \times 14.75 \times 10^{36}}{39.83 \times 10^{13}} = 7.33 \times 10^{22}\ \text{kg}$$
m₂ = 7.33 × 10²² kg, the Moon’s mass (7.34 × 10²² kg, the small difference coming from rounding).

$$d^2 = \frac{G m_1 m_2}{F} = \frac{6.67 \times 10^{-11} \times 5.972 \times 10^{24} \times 2 \times 10^{30}}{3.54 \times 10^{22}} = \frac{79.67 \times 10^{43}}{3.54 \times 10^{22}} = 2.25 \times 10^{22}$$
$$d = \sqrt{2.25 \times 10^{22}} = 1.5 \times 10^{11}\ \text{m}$$
d = 1.5 × 10¹¹ m (150 million km: the distance light covers in about 8⅓ minutes).

(The water is on the Earth, so we use the center-to-center distance.)
m₁ = 7.342 × 10²² kg, m₂ = 1 kg, d = 384 400 km = 3.844 × 10⁸ m.
$$F = \frac{6.67 \times 10^{-11} \times 7.342 \times 10^{22} \times 1}{(3.844 \times 10^{8})^2} = \frac{48.97 \times 10^{11}}{14.78 \times 10^{16}} = 3.314 \times 10^{-5}\ \text{N}$$
F = 3.31 × 10⁻⁵ N on every kilogram: about 300,000 times weaker than the Earth’s pull, yet over a whole ocean it is enough to raise the tides.

R = 1.737 × 10⁶ m.
$$g = \frac{GM}{R^2} = \frac{6.67 \times 10^{-11} \times 7.342 \times 10^{22}}{(1.737 \times 10^{6})^2} = \frac{48.97 \times 10^{11}}{3.017 \times 10^{12}} = 1.62\ \text{m/s}^2$$
g = 1.62 m/s².

The mass is the same; the radius changes from 6371 km to 3474 ÷ 2 = 1737 km.
Since g ∝ 1/R²:
$$\frac{g_\text{new}}{g} = \left(\frac{6371}{1737}\right)^2 = 3.668^2 = 13.45$$
g would be about 13.45 times as large: about 132 m/s².

R = 3.389 × 10⁶ m.
$$g = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{(3.389 \times 10^{6})^2} = \frac{42.69 \times 10^{12}}{11.49 \times 10^{12}} = 3.72\ \text{m/s}^2$$
W = mg = 200 × 3.72 = 743 N.

$$g_J = \frac{G (319 M)}{(11R)^2} = \frac{319}{121} \times \frac{GM}{R^2} = \frac{319}{121} \times 9.8 = 25.84\ \text{m/s}^2$$
W = mg = 100 × 25.84 = 2584 N.

d = R + h = 6371 km + 20 200 km = 26 571 km = 2.6571 × 10⁷ m.
$$g_1 = \frac{GM}{(R+h)^2} = \frac{6.67 \times 10^{-11} \times 5.972 \times 10^{24}}{(2.6571 \times 10^{7})^2} = \frac{39.83 \times 10^{13}}{7.060 \times 10^{14}} = 0.564\ \text{m/s}^2$$
g₁ = 0.56 m/s². The satellites that find your position on a phone map are in a place where g is only one seventeenth of what it is on the ground.

d = 6 371 000 + 8848.86 = 6 379 848.86 m = 6.3798 × 10⁶ m.
$$g_1 = \frac{6.67 \times 10^{-11} \times 5.972 \times 10^{24}}{(6.3798 \times 10^{6})^2} = \frac{39.83 \times 10^{13}}{40.70 \times 10^{12}} = 9.786\ \text{m/s}^2$$
W = mg₁ = 10 × 9.786 = 97.86 N. At sea level it would be 98.14 N.

She can exert the same force in both places, so the weight lifted is the same:
$$m \times 3.75 = 100 \times 9.8 \;\Rightarrow\; m = \frac{980}{3.75} = 261.33\ \text{kg}$$
m = 261.33 kg

u = 0, t = 2.5 s.
$$h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2} \times 9.8 \times 2.5^2 = 4.9 \times 6.25 = 30.63\ \text{m}$$
h = 30.6 m (ignoring the short time the sound takes to travel up).

u = 0, h = 15 m, g = 9.8 m/s².
Time: $h = \tfrac{1}{2}gt^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{30}{9.8}} = \sqrt{3.061} = 1.75\ \text{s}$
Velocity: $v = u + gt = 0 + 9.8 \times 1.75 = 17.15\ \text{m/s}$ (or $v = \sqrt{2gh} = \sqrt{294} = 17.15$ m/s).
t = 1.75 s, v = 17.15 m/s

u = 15 m/s; at the top v = 0; going up, g = −9.8 m/s².
$$v^2 = u^2 + 2gh \;\Rightarrow\; 0 = 15^2 + 2 \times (-9.8) \times h \;\Rightarrow\; h = \frac{225}{19.6} = 11.48\ \text{m}$$
h = 11.48 m.