11 Circular Motion and Gravitation Practice Problems

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. Centripetal acceleration from a speed and a radius

   A child rides near the edge of a playground roundabout, 2.4 m from the axis, and moves at a constant 3.1 m/s. (a) Find the magnitude and direction of the child's acceleration. (b) The roundabout is then spun so the child moves at 6.2 m/s at the same radius. Find the new acceleration without recomputing from scratch.

   Show the worked solution

   1. The speed is constant and the path is a circle, so the acceleration is purely centripetal. The sheet prints one line for it.

      $$a_c = \frac{v^2}{r} = \frac{(3.1\ \text{m/s})^2}{2.4\ \text{m}} = \frac{9.61}{2.4} = 4.004\ \text{m/s}^2$$

   2. Direction matters as much as the number. Essential knowledge 2.9.A.1.ii says centripetal acceleration is directed towards the centre of the object's circular path, so this acceleration points along the radius, inwards, and swings round to keep doing so as the child moves.

   3. Part (b). Doubling the speed at fixed radius multiplies $v^2$ by four, so the acceleration quadruples. No new arithmetic is needed.

      $$a_c = \frac{(6.2\ \text{m/s})^2}{2.4\ \text{m}} = \frac{38.44}{2.4} = 16.02\ \text{m/s}^2$$

   4. Two significant figures: $4.0\ \text{m/s}^2$ and $16\ \text{m/s}^2$.

   5. Note that the child's speed is not changing at all, yet the acceleration is nonzero. Acceleration is the rate of change of velocity, and velocity is a vector, so turning counts even at unchanging speed.

   (a) $a_c = 4.0\ \text{m/s}^2$, directed towards the axis of the roundabout. (b) $16\ \text{m/s}^2$, four times larger, because $a_c$ goes with the square of the speed. The child's mass was never given and was never needed: acceleration is set by the geometry and the speed alone.

2. 2. Working from a period instead of a speed

   A point on the rim of a bicycle wheel of radius 0.34 m completes one revolution every 0.52 s while the bike moves at a steady pace. Find the point's speed and its centripetal acceleration, and express the acceleration as a multiple of $g$.

   Show the worked solution

   1. One revolution covers one circumference, so the speed is the circumference divided by the period.

      $$v = \frac{2\pi r}{T} = \frac{2\pi (0.34\ \text{m})}{0.52\ \text{s}} = \frac{2.1363\ \text{m}}{0.52\ \text{s}} = 4.108\ \text{m/s}$$

   2. Now the centripetal acceleration from the printed relationship.

      $$a_c = \frac{v^2}{r} = \frac{(4.108\ \text{m/s})^2}{0.34\ \text{m}} = \frac{16.878}{0.34} = 49.64\ \text{m/s}^2$$

   3. As a check, combine the two steps symbolically before substituting. Putting $v = 2\pi r/T$ into $a_c = v^2/r$ gives a form that takes the period directly:

      $$a_c = \frac{4\pi^2 r}{T^2} = \frac{4\pi^2 (0.34\ \text{m})}{(0.52\ \text{s})^2} = \frac{13.423}{0.2704} = 49.64\ \text{m/s}^2$$

      Same answer, and it avoids carrying a rounded speed through.

   4. Compare with gravity: $49.64 / 9.8 = 5.07$, so the rim point accelerates at about 5.1 times $g$.

   5. Two significant figures: $v = 4.1\ \text{m/s}$ and $a_c = 50\ \text{m/s}^2$.

   $v = 4.1\ \text{m/s}$ and $a_c = 50\ \text{m/s}^2$ towards the hub, about 5.1 times $g$. Small radii produce large accelerations at ordinary speeds, which is why a rim magnet needs a firm mount and why the same 4.1 m/s on a circle of radius 100 m would give only $0.17\ \text{m/s}^2$.

3. 3. A puck on a string: a horizontal circle

   A 0.24 kg puck slides on a frictionless horizontal air table, tied to a light string of length 0.65 m whose other end is pinned to the table. The puck travels in a circle at a constant 3.8 m/s. (a) Find the centripetal acceleration and the tension in the string. (b) The string is rated to 40 N. Find the fastest the puck could travel before it snaps.

   Show the worked solution

   1. Draw the real forces first. Three act on the puck: the weight down, the normal force from the table up, and the string tension pulling horizontally towards the pin. The table is horizontal, so the vertical pair cancel exactly, and the puck's weight of $(0.24)(9.8) = 2.35\ \text{N}$ plays no part in what follows.

   2. That leaves tension as the only horizontal force. Take positive as pointing towards the centre of the circle. Newton's second law along the radius then reads

      $$T = m a_c$$

      There is no extra centripetal force here. The tension is the net radial force, and $ma_c$ is what it has to equal.

   3. Part (a). Compute the acceleration first.

      $$a_c = \frac{v^2}{r} = \frac{(3.8\ \text{m/s})^2}{0.65\ \text{m}} = \frac{14.44}{0.65} = 22.22\ \text{m/s}^2$$
      $$T = (0.24\ \text{kg})(22.22\ \text{m/s}^2) = 5.332\ \text{N}$$

   4. Part (b). Set the tension to the string's limit and solve backwards for the speed.

      $$T_{\text{max}} = \frac{m v_{\text{max}}^2}{r} \quad \Rightarrow \quad v_{\text{max}} = \sqrt{\frac{T_{\text{max}}\, r}{m}}$$
      $$v_{\text{max}} = \sqrt{\frac{(40\ \text{N})(0.65\ \text{m})}{0.24\ \text{kg}}} = \sqrt{\frac{26}{0.24}} = \sqrt{108.33} = 10.41\ \text{m/s}$$

   5. Check it: at 10.41 m/s the acceleration is $108.33/0.65 = 166.7\ \text{m/s}^2$ and the required force is $(0.24)(166.7) = 40.0\ \text{N}$, right at the limit.

   6. Two significant figures: $a_c = 22\ \text{m/s}^2$, $T = 5.3\ \text{N}$, $v_{\text{max}} = 10\ \text{m/s}$.

   (a) $a_c = 22\ \text{m/s}^2$ and $T = 5.3\ \text{N}$. (b) The string snaps above $10\ \text{m/s}$. The moment it snaps the tension is gone, so the net force on the puck drops to zero and it slides off along the tangent in a straight line at constant speed, not outwards along the radius.

4. 4. A car on a flat curve: friction supplies the net inward force

   A 1350 kg car rounds a flat, unbanked curve of radius 68 m at a constant 21 m/s. (a) Find the centripetal acceleration and the size of the net inward force needed, and name the force that supplies it. (b) Find the smallest coefficient of static friction that makes this possible. (c) If the road is wet and the coefficient of static friction is only 0.45, find the fastest the car could take the same curve.

   Show the worked solution

   1. Real forces on the car: the weight down, the normal force up, and friction from the road, horizontal. The road is flat, so the vertical forces cancel and $F_N = mg$. Nothing else touches the car, so friction is the only horizontal force and therefore the only candidate to supply the net inward force. Take positive as pointing towards the centre of the curve.

   2. Part (a). The acceleration first.

      $$a_c = \frac{v^2}{r} = \frac{(21\ \text{m/s})^2}{68\ \text{m}} = \frac{441}{68} = 6.485\ \text{m/s}^2$$
      $$F_{\text{net, inward}} = m a_c = (1350\ \text{kg})(6.485\ \text{m/s}^2) = 8755\ \text{N}$$

      That 8755 N is not an extra force on the diagram. It is the value that static friction must take.

   3. Part (b). Static friction can supply at most $\mu_s F_N = \mu_s mg$. Set that ceiling equal to what is required.

      $$\mu_s\, mg \geq m a_c \quad \Rightarrow \quad \mu_s \geq \frac{a_c}{g} = \frac{6.485}{9.8} = 0.662$$

      The mass cancels, so a lorry and a motorbike need the same coefficient on the same curve at the same speed.

   4. Part (c). Rearrange the same inequality for speed, with the coefficient now fixed at 0.45.

      $$\frac{mv^2}{r} = \mu_s mg \quad \Rightarrow \quad v_{\text{max}} = \sqrt{\mu_s g r}$$
      $$v_{\text{max}} = \sqrt{(0.45)(9.8\ \text{m/s}^2)(68\ \text{m})} = \sqrt{299.9} = 17.32\ \text{m/s}$$

   5. Two significant figures: $a_c = 6.5\ \text{m/s}^2$, $F = 8.8 \times 10^3\ \text{N}$, $\mu_s \geq 0.66$, and $v_{\text{max}} = 17\ \text{m/s}$.

   6. Note that friction here is static, not kinetic, even though the car is moving fast. The relevant question is whether the tyre surface in contact is sliding across the road, and on a car that is cornering without skidding it is not.

   (a) $a_c = 6.5\ \text{m/s}^2$ and the net inward force is $8.8 \times 10^3\ \text{N}$, supplied entirely by static friction between the tyres and the road. (b) $\mu_s$ must be at least $0.66$, which a dry road can usually manage but a wet one cannot. (c) On a wet road with $\mu_s = 0.45$ the limit drops to $17\ \text{m/s}$. Exceed it and the required inward force outruns what friction can give, and the car runs wide.

5. 5. A vertical circle: tension at the bottom and at the top

   A 0.36 kg ball is swung in a vertical circle on a light string of length 0.95 m. Find the tension in the string (a) at the lowest point, where the ball is moving at 5.2 m/s, and (b) at the highest point, where it is moving at 3.4 m/s. Compare both with the ball's weight.

   Show the worked solution

   1. Two real forces act on the ball everywhere on the circle: the string tension along the string, and the weight straight down. What changes between top and bottom is not the forces but their directions relative to the centre. Compute the weight once.

      $$mg = (0.36\ \text{kg})(9.8\ \text{m/s}^2) = 3.528\ \text{N}$$

   2. Part (a), the lowest point. The centre of the circle is directly above the ball, so take positive as up. Tension points up towards the centre, so it is positive; the weight points down, away from the centre, so it is negative.

      $$T - mg = \frac{mv^2}{r}$$

   3. Substitute. The radial acceleration at the bottom is $a_c = (5.2)^2/0.95 = 27.04/0.95 = 28.46\ \text{m/s}^2$.

      $$T = m\left(\frac{v^2}{r} + g\right) = (0.36)(28.46 + 9.8) = (0.36)(38.26) = 13.77\ \text{N}$$

   4. Part (b), the highest point. Now the centre is directly below the ball, so positive points down. Both the tension and the weight point downwards, towards the centre, so both are positive and they share the work.

      $$T + mg = \frac{mv^2}{r}$$

   5. The radial acceleration at the top is $a_c = (3.4)^2/0.95 = 11.56/0.95 = 12.17\ \text{m/s}^2$.

      $$T = m\left(\frac{v^2}{r} - g\right) = (0.36)(12.17 - 9.8) = (0.36)(2.368) = 0.8526\ \text{N}$$

   6. Two significant figures: $T_{\text{bottom}} = 14\ \text{N}$ and $T_{\text{top}} = 0.85\ \text{N}$. Against the 3.53 N weight, that is about 3.9 times the weight at the bottom and about 0.24 times it at the top.

   7. Sanity check on the top value. Gravity alone could supply an inward force of 3.53 N, which corresponds to a speed of $\sqrt{gr} = \sqrt{(9.8)(0.95)} = 3.05\ \text{m/s}$. The ball is going faster than that, at 3.4 m/s, so it needs more inward force than gravity provides and the string must be taut. A positive tension was the expected result.

   $T = 14\ \text{N}$ at the bottom and $0.85\ \text{N}$ at the top, a factor of 16 apart with the same string, the same ball and similar speeds. At the bottom the tension has to beat gravity and turn the ball; at the top gravity is already pulling towards the centre and does most of the turning for free. That asymmetry is why a string breaks at the bottom of a swing.

6. 6. A loop: minimum speed at the top, apparent weight at the bottom

   A roller coaster car runs around a vertical circular loop of radius 8.5 m. (a) Find the minimum speed the car can have at the top of the loop and still stay on the track. (b) The car, loaded to 480 kg, passes the lowest point of the loop at 18 m/s. Find the normal force the track exerts on it there, and compare that with the car's weight.

   Show the worked solution

   1. Part (a). At the top, both the weight and the track's normal force point downwards, which is towards the centre. Take positive as downwards there.

      $$F_N + mg = \frac{mv^2}{r}$$

      The track can push but never pull, so $F_N \geq 0$. The slowest possible case is $F_N = 0$, where gravity alone causes the centripetal acceleration. That is exactly what essential knowledge 2.9.A.2.i describes.

   2. Set $F_N = 0$ and the mass cancels, giving the CED's derived equation.

      $$mg = \frac{mv^2}{r} \quad \Rightarrow \quad v_{\text{min}} = \sqrt{gr} = \sqrt{(9.8\ \text{m/s}^2)(8.5\ \text{m})} = \sqrt{83.3} = 9.127\ \text{m/s}$$

   3. Any slower and the required inward force $mv^2/r$ would be less than the weight. The track cannot pull outwards to make up the difference, so the car would leave the rails before reaching the top. The result contains no mass, so it holds for an empty car and a full one alike.

   4. Part (b). At the bottom the centre of the loop is above the car, so take positive as up. The normal force points up towards the centre and the weight points down.

      $$F_N - mg = \frac{mv^2}{r}$$

   5. The radial acceleration is $a_c = (18)^2/8.5 = 324/8.5 = 38.12\ \text{m/s}^2$.

      $$F_N = m\left(\frac{v^2}{r} + g\right) = (480\ \text{kg})(38.12 + 9.8) = (480)(47.92) = 23{,}000\ \text{N}$$

   6. The car's weight is $(480)(9.8) = 4704\ \text{N}$, so the normal force is $23{,}000/4704 = 4.89$ times the weight. Because apparent weight is the magnitude of the normal force (essential knowledge 2.6.C.1), a rider at the bottom of this loop feels roughly 4.9 times their usual weight.

   7. Two significant figures: $v_{\text{min}} = 9.1\ \text{m/s}$ and $F_N = 2.3 \times 10^4\ \text{N}$.

   (a) $v_{\text{min}} = 9.1\ \text{m/s}$ at the top, independent of the car's mass. (b) $F_N = 2.3 \times 10^4\ \text{N}$ at the bottom, about 4.9 times the car's 4704 N weight. The same rider is pressed hard into the seat at the bottom and nearly weightless at the top, from a track that only ever pushes.

7. 7. A banked curve with no friction needed

   A highway curve of radius 95 m is to be banked so that a car can round it at 24 m/s with no friction at all between the tyres and the road. (a) Find the required banking angle. (b) For a 1150 kg car travelling at exactly the design speed, find the normal force from the road. (c) Say qualitatively what happens if a car takes the same bank at 30 m/s.

   Show the worked solution

   1. With no friction, only two real forces act on the car: the weight straight down and the normal force perpendicular to the road surface, which is tilted by the banking angle $\theta$. The car moves in a horizontal circle, so its acceleration is horizontal and points towards the centre of the curve. Take positive $x$ as horizontal towards the centre and positive $y$ as up.

   2. Vertical axis. There is no vertical acceleration, so the vertical component of the normal force carries the whole weight.

      $$F_N\cos\theta - mg = 0 \quad \Rightarrow \quad F_N\cos\theta = mg$$

   3. Horizontal axis. The horizontal component of the normal force is the only force pointing towards the centre, so it is the net inward force.

      $$F_N\sin\theta = \frac{mv^2}{r}$$

   4. Part (a). Divide the second equation by the first. Both $F_N$ and $m$ cancel, which is why the design angle does not depend on the vehicle.

      $$\tan\theta = \frac{v^2}{rg} = \frac{(24\ \text{m/s})^2}{(95\ \text{m})(9.8\ \text{m/s}^2)} = \frac{576}{931} = 0.6187$$
      $$\theta = \arctan(0.6187) = 31.74^\circ$$

   5. Part (b). Rather than divide by a rounded cosine, square and add the two axis equations, which gives the normal force directly from its two components.

      $$F_N = \sqrt{(mg)^2 + \left(\frac{mv^2}{r}\right)^2}$$

      Here $mg = (1150)(9.8) = 11{,}270\ \text{N}$ and $mv^2/r = (1150)(576)/95 = 6973\ \text{N}$, so

      $$F_N = \sqrt{(11{,}270)^2 + (6973)^2} = \sqrt{1.7563 \times 10^8} = 13{,}253\ \text{N}$$

      That is about 1.18 times the car's weight, which is the price of tilting the support.

   6. Part (c). At 30 m/s the required inward force is larger than the bank alone can provide, so the car tends to slide up and out of the curve, and friction must act down the slope to make up the difference. The CED keeps this case qualitative: the Topic 2.9 boundary statement says AP Physics 1 only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion, and that analysis of situations in which friction is required on a banked curve is limited to qualitative descriptions.

   7. Two significant figures: $\theta = 32^\circ$ and $F_N = 1.3 \times 10^4\ \text{N}$.

   (a) $\theta = 32^\circ$, independent of the car's mass. (b) $F_N = 1.3 \times 10^4\ \text{N}$, about 18 percent above the car's 11,270 N weight. (c) Above the design speed the bank alone is not enough, so friction has to act down the slope, and AP Physics 1 asks you to describe that rather than compute it. The design speed is the one speed at which a perfectly icy bank would still work.

8. 8. A conical pendulum

   A 0.65 kg ball hangs from a light string 1.20 m long and is swung so that it travels in a horizontal circle with the string making a constant angle of $24^\circ$ with the vertical. Find the tension in the string, the radius of the circle, the ball's speed, and the time for one revolution.

   Show the worked solution

   1. Only two real forces act on the ball: the string tension along the string, tilted $24^\circ$ from vertical, and the weight straight down. Essential knowledge 2.9.A.2.iii names this case directly: a component of tension contributes to the net force producing the centripetal acceleration for a conical pendulum. Take positive $x$ as horizontal towards the axis of the circle and positive $y$ as up.

   2. Vertical axis. The ball stays at the same height, so there is no vertical acceleration and the vertical component of the tension must carry the whole weight.

      $$T\cos 24^\circ = mg = (0.65\ \text{kg})(9.8\ \text{m/s}^2) = 6.37\ \text{N}$$
      $$T = \frac{6.37\ \text{N}}{0.91355} = 6.973\ \text{N}$$

      The tension exceeds the 6.37 N weight, because only part of it is pointing upwards.

   3. The radius is not the string length. It is the horizontal distance from the ball to the axis.

      $$r = L\sin 24^\circ = (1.20\ \text{m})(0.40674) = 0.4881\ \text{m}$$

   4. Horizontal axis. The horizontal component of the tension is the only force pointing towards the axis, so it is the net inward force.

      $$T\sin 24^\circ = \frac{mv^2}{r}$$
      $$T\sin 24^\circ = (6.973\ \text{N})(0.40674) = 2.836\ \text{N}$$

   5. Solve for the speed.

      $$v = \sqrt{\frac{(2.836\ \text{N})(0.4881\ \text{m})}{0.65\ \text{kg}}} = \sqrt{2.130} = 1.459\ \text{m/s}$$

   6. Cross-check by dividing the horizontal equation by the vertical one, which cancels both $T$ and $m$ and gives a formula with no tension in it:

      $$\tan\theta = \frac{v^2}{rg} \quad \Rightarrow \quad v = \sqrt{g r \tan\theta} = \sqrt{(9.8)(0.4881)(0.44523)} = \sqrt{2.130} = 1.459\ \text{m/s}$$

      Same value, and notice it is the identical relationship the banked curve produced in problem 7, for the same reason: a tilted force supplying an inward component while holding up a weight.

   7. The period is one circumference at that speed.

      $$T_{\text{period}} = \frac{2\pi r}{v} = \frac{2\pi(0.4881\ \text{m})}{1.459\ \text{m/s}} = \frac{3.0667}{1.459} = 2.101\ \text{s}$$

   8. Two significant figures: tension $7.0\ \text{N}$, radius $0.49\ \text{m}$, speed $1.5\ \text{m/s}$, period $2.1\ \text{s}$.

   $T = 7.0\ \text{N}$, $r = 0.49\ \text{m}$, $v = 1.5\ \text{m/s}$, and one revolution takes $2.1\ \text{s}$. The two most common errors here are using the string length 1.20 m as the radius, which inflates the speed, and writing $T = mg$, which ignores that the string is tilted and cannot hold the ball up with its full magnitude.

9. 9. Newton's law of universal gravitation on another world

   A planet has mass $4.9 \times 10^{24}\ \text{kg}$ and radius $5.4 \times 10^{6}\ \text{m}$. Using $G = 6.67 \times 10^{-11}\ \text{N} \cdot \text{m}^2/\text{kg}^2$, find the gravitational field strength at its surface, and the weight there of a 72 kg astronaut. Compare that weight with her weight on Earth.

   Show the worked solution

   1. Start from the printed law of universal gravitation for the force between the planet and an object of mass $m$ resting on its surface, a distance $R$ from the planet's centre.

      $$\left|\vec{F}_g\right| = \frac{GMm}{R^2}$$

   2. Field strength is force per unit mass, which is the CED's derived equation under 2.6.A.2.i. Dividing by $m$ removes the astronaut from the question entirely.

      $$g_{\text{planet}} = \frac{GM}{R^2}$$

   3. Substitute, doing the powers of ten separately so they can be checked.

      $$GM = (6.67 \times 10^{-11})(4.9 \times 10^{24}) = 3.268 \times 10^{14}$$
      $$R^2 = (5.4 \times 10^{6})^2 = 2.916 \times 10^{13}\ \text{m}^2$$
      $$g_{\text{planet}} = \frac{3.268 \times 10^{14}}{2.916 \times 10^{13}} = 11.21\ \text{m/s}^2$$

   4. Now the astronaut's weight there, which is just $mg$ with the local $g$ (the CED's derived equation under 2.6.A.3).

      $$F_g = (72\ \text{kg})(11.21\ \text{m/s}^2) = 807.1\ \text{N}$$

   5. On Earth the same astronaut weighs $(72\ \text{kg})(9.8\ \text{m/s}^2) = 705.6\ \text{N}$, so she is about 14 percent heavier on this planet.

      $$\frac{807.1}{705.6} = 1.14$$

   6. Two significant figures: $g_{\text{planet}} = 11\ \text{m/s}^2$ and a weight of $8.1 \times 10^2\ \text{N}$ against $7.1 \times 10^2\ \text{N}$ on Earth.

   7. Worth noticing: this planet has only 82 percent of Earth's mass yet a stronger surface field, because its radius is 15 percent smaller and the radius is squared in the denominator. Surface gravity is not a mass ranking.

   $g_{\text{planet}} = 11\ \text{m/s}^2$ and the astronaut weighs $8.1 \times 10^2\ \text{N}$ there, about 14 percent more than her $7.1 \times 10^2\ \text{N}$ on Earth. Her mass is 72 kg in both places and did not change. Only the field she is standing in did.

10. 10. A satellite in a circular orbit

    A satellite orbits Earth in a circular path 1200 km above the surface. Take Earth's mass as $5.97 \times 10^{24}\ \text{kg}$, its radius as $6.37 \times 10^{6}\ \text{m}$, and $G = 6.67 \times 10^{-11}\ \text{N} \cdot \text{m}^2/\text{kg}^2$. Find the satellite's orbital speed and its period. Then state, without recomputing, how each would change if the orbital radius were doubled.

    Show the worked solution

    1. First get the orbital radius, measured from Earth's centre, not the surface. The 1200 km altitude has to be converted and added.

       $$r = 6.37 \times 10^{6}\ \text{m} + 1.20 \times 10^{6}\ \text{m} = 7.57 \times 10^{6}\ \text{m}$$

    2. Only one real force acts on the satellite: Earth's gravitational pull, directed towards Earth's centre. Essential knowledge 2.9.B.1 says exactly this, that for a satellite in circular orbit the centripetal acceleration is caused only by gravitational attraction. Take positive as pointing towards Earth's centre and write Newton's second law along the radius.

       $$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

    3. The satellite's own mass $m$ appears on both sides and cancels, which is why a bolt and a space station in the same orbit travel at the same speed. One factor of $r$ cancels too.

       $$v = \sqrt{\frac{GM}{r}}$$

    4. Substitute.

       $$GM = (6.67 \times 10^{-11})(5.97 \times 10^{24}) = 3.982 \times 10^{14}$$
       $$v = \sqrt{\frac{3.982 \times 10^{14}}{7.57 \times 10^{6}}} = \sqrt{5.260 \times 10^{7}} = 7.253 \times 10^{3}\ \text{m/s}$$

    5. The period is one circumference at that speed.

       $$T = \frac{2\pi r}{v} = \frac{2\pi(7.57 \times 10^{6}\ \text{m})}{7.253 \times 10^{3}\ \text{m/s}} = \frac{4.756 \times 10^{7}}{7.253 \times 10^{3}} = 6.558 \times 10^{3}\ \text{s}$$

       That is 109 minutes, a shade under two hours.

    6. The scaling question needs no new arithmetic. Since $v \propto r^{-1/2}$, doubling the radius multiplies the speed by $1/\sqrt{2} = 0.707$. And since $T = 2\pi r / v \propto r^{3/2}$, doubling the radius multiplies the period by $2^{3/2} = 2.83$. Higher orbits are slower and longer, which is Kepler's third law in the form Topic 2.9's learning objective 2.9.B asks about.

    7. Three significant figures, matching the data: $v = 7.25 \times 10^{3}\ \text{m/s}$ and $T = 6.56 \times 10^{3}\ \text{s}$.

    $v = 7.25 \times 10^{3}\ \text{m/s}$, about 7.25 km/s, and $T = 6.56 \times 10^{3}\ \text{s}$, about 109 minutes. Doubling the radius would cut the speed to 0.707 of that and stretch the period by a factor of 2.83. The satellite's own mass cancelled and never mattered. The single most common error in this problem is using the 1200 km altitude as $r$ instead of adding it to Earth's radius.

11. 11. Explain why: there is no centrifugal force on the free-body diagram

    A passenger sitting in the back of a car slides towards the right-hand door as the car takes a sharp left turn at constant speed. She says a centrifugal force pushed her outwards. Explain what actually happened, say what force acts on her and in which direction, and explain why nothing is drawn pointing away from the centre on her free-body diagram.

    Show the worked solution

    1. Start with what a force is. Essential knowledge 2.2.A.1.i says a force exerted on an object is always due to the interaction of that object with another object. So the test for any proposed force is a single question: what object is exerting it? For the outward push there is no answer. No object is on her left pushing her to the right.

    2. Now describe the motion honestly, from the ground. Before the turn she was moving in a straight line. Newton's first law says her velocity will keep on in that straight line unless a net force acts (essential knowledge 2.4.A.3). The car turns left; she initially does not, because nothing has yet acted on her sideways. The gap between her straight path and the car's curved path closes until the door reaches her.

    3. Then a real interaction begins. The door pushes on her, and it pushes inwards, towards the centre of the turn, because that is the only way a surface on her right can push. That inward normal force is the net radial force, and it is what makes her turn with the car.

       $$F_{\text{door on passenger}} = \frac{mv^2}{r} \quad \text{towards the centre of the turn}$$

    4. So why does it feel outward? Because she is judging from inside the car, which is an accelerating and therefore noninertial reference frame. Essential knowledge 2.4.A.5 defines an inertial reference frame as one from which an observer would verify Newton's first law, and a turning car is not one. The sensation is her body's inertia resisting a real inward push, not evidence of an outward force.

    5. Her free-body diagram, drawn in the ground frame, has three arrows and no more: her weight down, the seat's normal force up, and the door's normal force pointing horizontally towards the centre of the turn. Essential knowledge 2.2.B.2 says the diagram shows each of the forces exerted on the object by the environment, and no environment object is pushing outwards.

    6. The final point is the one that generalises to everything else on this page. Centripetal is a direction, not a force. You never add an $F_c$ arrow to a diagram. You draw the real forces, resolve them along the radius, and set their inward sum equal to $mv^2/r$. Whether that sum comes from tension (problem 3), friction (problem 4), tension and gravity together (problem 5), a normal force (problem 6), a tilted normal force (problem 7), or gravitational attraction (problem 10), the procedure never changes.

    There is no centrifugal force, because no object exerts it, and essential knowledge 2.2.A.1.i says every force on an object comes from an interaction with another object. What happened is that she continued in a straight line while the car curved away beneath her, exactly as Newton's first law predicts, until the door reached her and pushed her inwards. Her free-body diagram carries her weight, the seat's normal force, and that inward push from the door: nothing points outwards. The outward feeling is what inertia feels like from inside a noninertial frame.