Earth and the Solar System Worked ExamplesIGCSE Physics 0625

The Sun does not move across our sky itself. The Earth spins on its axis once in about $24$ hours, carrying the observer from the sunlit side into shadow and back. That causes the apparent daily motion of the Sun and the cycle of day and night. So the answer is B.

A is wrong because the Earth takes about $365$ days to orbit the Sun (that gives the seasons, not the day). C is wrong because the Sun does not orbit the Earth. D is wrong because the Moon takes about one month to orbit the Earth (that gives the cycle of the Moon's phases).

(a) P to Q is one quarter of a complete orbit, and the Earth takes about $365$ days to orbit the Sun once.

(b) The north end of the axis tilts to the right at every position. At R (to the right of the Sun) the north end tilts away from the Sun, so the southern hemisphere tilts towards it. So it is summer in the southern hemisphere at R (and winter in the northern hemisphere).

Check with P: there the north end tilts towards the Sun, so P is northern summer, half an orbit (about six months) from R, as expected.

(c) Because the axis always points the same way in space, each hemisphere tilts towards the Sun and then away from it once in every orbit. The Earth orbits the Sun once in about $365$ days, so the seasons repeat every year.

(a) $r$ is the average radius of the orbit (the average distance of the planet from the Sun). $T$ is the orbital period: the time taken to complete one orbit of the Sun.

(b) The speed is needed in m/s, so convert the period to seconds first.

Substitute into the equation.

Evaluate and give the unit.

Start from the equation, written out, and rearrange it for $T$.

Convert the speed to m/s: $6.80 \text{ km/s} = 6800$ m/s. Substitute.

Work out the number of seconds in one Earth year ($365$ days).

Divide, keeping the unrounded value of $T$ until the end.

The orbital radius is measured from the centre of the Earth, so add the Earth's radius to the height.

Convert the period to seconds.

Use the orbital speed equation.

So the answer is C. B comes from using the height alone as $r$. A comes from using $24$ hours for $T$. D comes from using $1$ hour for $T$.

(a) Saturn: its density ($0.7$ g/cm³) is far lower than the others (about $4$–$5$ g/cm³), which suggests it is made mostly of gas, while the others are rocky.

(b) The greater the orbital distance, the longer the orbital duration, and the duration increases by a larger factor than the distance (Saturn is about $25$ times further away than Mercury but takes over $120$ times as long).

(c) Subtract $273$ from the kelvin temperature.

(d) On Venus. Its gravitational field strength ($8.9$ N/kg) is greater than on Mars ($3.7$ N/kg), so the gravitational force on the ball is greater and its acceleration is greater ($a = g$ with no air resistance). It therefore covers the same distance in less time.

(e) Venus ($740$ K) is hotter than Mercury ($440$ K), even though Mercury is closer to the Sun.

(a) Convert kilometres to metres first.

Use time = distance ÷ speed.

(b) The command goes there and the reply comes back, so double the time, then convert to minutes.

(c) At Y: $4.0 \times 10^{8}$ km $= 4.0 \times 10^{11}$ m.

(d) Earth and Mars both orbit the Sun but at different distances and with different orbital periods (Mars takes longer), so their positions relative to each other change continuously: sometimes on the same side of the Sun, sometimes on opposite sides.

(a) Gravitational field strength is the force per unit mass acting on a mass in the field ($g = W/m$).

(b) The mass of the planet (a more massive planet has a stronger field at its surface).

(c) Venus has the greater orbital speed. Marking points: Venus is closer to the Sun; the strength of the Sun's gravitational field decreases with distance from the Sun; so the orbital speeds of planets decrease as their distance from the Sun increases.

(a) Fastest at A, the point closest to the Sun. Slowest at C, the point furthest from the Sun.

(b) From C to A the dwarf planet moves closer to the Sun, so its gravitational potential energy decreases.

Energy is conserved (no energy is lost in space), so the decrease in gravitational potential energy equals the increase in kinetic energy. More kinetic energy means greater speed.

(c) The Sun is not at the centre of the ellipse. It is offset towards one end (nearer A). It would be at the centre only if the orbit were approximately circular.

(a) At the surface (distance $5.0 \times 10^{6}$ m) the graph gives $g = 8.0$ N/kg.

(b) The graph uses distance from the centre. Height above the surface plus the radius:

Read up from $10$ on the distance axis: $g = 2.0$ N/kg.

(c) Mass is the amount of matter and does not depend on the field.

(d) $g$ decreases as the distance increases. It falls quickly at first and then more gradually, getting closer and closer to zero.

(a) Rearrange the orbital speed equation for $r$.

Use SI units: $v = 3.50 \times 10^{4}$ m/s and $T = 225 \times 24 \times 3600 = 1.944 \times 10^{7}$ s.

(b) On the same side of the Sun, the distance between the planets is the difference between their orbital radii.

The pulse travels there and back, so the total distance is $2d$.

(a) Convert the speed to m/s: $54.5$ km/s $= 5.45 \times 10^{4}$ m/s.

(b) Kinetic energy at the furthest point, with $v = 910$ m/s:

Subtract. The far-point value is tiny in comparison, so almost all the kinetic energy is lost.

(c) As the comet moves away from the Sun it does work against the Sun's gravitational attraction: the kinetic energy is transferred to its gravitational potential energy store. On the way back towards the Sun, gravitational potential energy decreases and, because energy is conserved, kinetic energy increases by the same amount, so the comet speeds up.

(d) The Sun is not at the centre of the ellipse. It is much closer to one end of the orbit, the end where the comet passes nearest.