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How do we describe projectile motion and the gravitational force between masses?

Projectile motion resolved into independent horizontal and vertical components, Newton's law of universal gravitation, and the gravitational field around a mass.

An SQA Higher Physics answer on gravitation, covering projectile motion resolved into independent horizontal and vertical components, Newton's law of universal gravitation, and the idea of a gravitational field around a mass.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. Projectile motion
  3. Newton's law of universal gravitation
  4. The gravitational field
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to treat projectile motion as two independent motions (constant-velocity horizontal and accelerating vertical), apply Newton's law of universal gravitation to find the force between two masses, and describe the gravitational field around a mass.

Projectile motion

To launch at an angle θ\theta with speed vv, resolve into vh=vcosθv_h = v\cos\theta and vv=vsinθv_v = v\sin\theta, then apply the equations of motion to each component. The horizontal motion uses dh=vhtd_h = v_h t; the vertical motion uses v=u+atv = u + at and s=ut+12at2s = ut + \tfrac{1}{2}at^2 with a=ga = -g. This independence is why a bullet fired horizontally and one dropped from the same height hit the ground at the same instant.

Newton's law of universal gravitation

Every mass attracts every other mass with a force proportional to the product of the masses and inversely proportional to the square of their separation. Because GG is tiny, the force between everyday objects is negligible; it becomes important only when at least one mass is astronomical, such as a planet or star. Doubling the separation quarters the force (the inverse-square dependence).

The gravitational field

This is why gg at the Earth's surface is about 9.8 N kg19.8\ \text{N kg}^{-1} but decreases with height, and why the Moon (smaller mass) has a surface field of only about 1.6 N kg11.6\ \text{N kg}^{-1}. A field line diagram for a planet shows lines pointing radially inward, closer together near the surface where the field is stronger.

Examples in context

A long jumper or a shot put follows a parabolic path set by the independent horizontal and vertical motions; the optimum launch angle near 4545^{\circ} balances range against height. Satellites and the International Space Station stay in orbit because the gravitational force GMmr2\frac{GMm}{r^2} provides exactly the centripetal force needed; astronauts feel weightless because they and the station fall together at the same gg. Tides arise from the difference in the Moon's gravitational pull across the Earth. Spacecraft trajectories to other planets are computed from Newton's law of gravitation, using the inverse-square weakening of each body's field to plan slingshot manoeuvres.

Try this

Q1. State what happens to the horizontal velocity of a projectile during flight when air resistance is ignored. [1 mark]

  • Cue. It stays constant (there is no horizontal force).

Q2. A stone is dropped from rest off a bridge and takes 2.0 s2.0\ \text{s} to reach the water. Calculate the height of the bridge. Take g=9.8 m s2g = 9.8\ \text{m s}^{-2}. [2 marks]

  • Cue. h=12gt2=12×9.8×2.02=19.6 mh = \tfrac{1}{2}gt^2 = \tfrac{1}{2} \times 9.8 \times 2.0^2 = 19.6\ \text{m}.

Q3. State how the gravitational force between two masses changes if the distance between them is doubled. [1 mark]

  • Cue. It becomes one quarter of its original value (inverse-square law).

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20185 marksA ball is launched horizontally from the top of a cliff at 8.0 m per second and lands 24 m from the base of the cliff. Calculate the time of flight and the height of the cliff. Take g as 9.8 m per second squared and ignore air resistance.
Show worked answer →

Horizontal and vertical motions are independent.

Time of flight from the horizontal motion (constant velocity). Relationship: dh=vhtd_h = v_h t. Substitution: 24=8.0×t24 = 8.0 \times t. Answer: t=3.0t = 3.0 s.

Height from the vertical motion (free fall from rest). Relationship: h=12gt2h = \tfrac{1}{2}g t^2. Substitution: h=12×9.8×3.02h = \tfrac{1}{2} \times 9.8 \times 3.0^2. Answer: h=44.1h = 44.1 m.

Markers reward using the horizontal motion to find the time, then the vertical motion with that time, and both answers with units. A common error is mixing the horizontal speed into the vertical equation.

SQA Higher 20213 marksCalculate the gravitational force of attraction between two 5.0 kg masses whose centres are 0.40 m apart. Take the gravitational constant G as 6.67 times ten to the power minus eleven N metre squared per kilogram squared.
Show worked answer →

Use Newton's law of universal gravitation.

Relationship: F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}.

Substitution: F=6.67×1011×5.0×5.00.402F = \frac{6.67 \times 10^{-11} \times 5.0 \times 5.0}{0.40^2}.

Answer: F=6.67×1011×250.16=1.04×108F = \frac{6.67 \times 10^{-11} \times 25}{0.16} = 1.04 \times 10^{-8} N.

Markers reward the correct relationship, squaring the separation in the denominator, and the very small answer with unit, showing why gravity between everyday masses is negligible.

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