IPhO 2022, Експеримент — Задача 1. Planet

• A.1 15.7 m/s²

  Official value g = 15.7 m/s² with Δg = 0.5 m/s²; the marking scheme gives full marks for 15.0–16.4 m/s².

• A.2 $R = \frac{L^2 - H^2}{2H}$ ≈ $13200$ km

• A.3 $4.2 \times 10^{25}$ kg

  Official: M = gR²/G = 4.2 × 10²⁵ kg with ΔM = 0.2 × 10²⁵ kg; the effect is the centrifugal force due to the planet's rotation, which reduces the measured acceleration and hence the estimate of M.

• B.1 1.31 m/s

  Official value u = 1.31 m/s with Δu = 0.04 m/s.

• B.2 0.6 kg/m³

  Official value ρ_a0 = 0.60 kg/m³ with Δρ_a0 = 0.07 kg/m³.

• B.3 7500 m

  Official value H₀ = 7500 m with ΔH₀ ≈ 2000 m.

• B.4 μ = 72 g/mol ≈ 70 g/mol with Δμ = 20 g/mol; p₀ = 20 000 Pa with Δp₀ = 6000 Pa.

• C.1 28000 s

  Official value T = 28 000 s ≈ 8 h with ΔT = 0.2 h.

E1: Planet - SOLUTION

A.1: The free-fall acceleration $g$ can be found by dropping the ball from low heights such that the air friction and effects from the curvature of the planet are minimized. We also choose the radius and density of the ball to be as big as possible to minimize the effect of air friction, i.e. setting $r = 50\ \mathrm{cm}$, $\rho = 10\ \mathrm{g/cm^3}$. The drop height is then given by $h = gt^2/2$, and so we can find $g$ from the slope of $t^2$ vs $h$. From the graph, we measure the slope $2/g = 0.127\ \mathrm{s^2/m}$ and its error $\Delta(2/g) = 0.004\ \mathrm{s^2/m}$ and so $g = 15.7\ \mathrm{m/s^2}$ with an error of $\Delta g = 0.5\ \mathrm{m/s^2}$.

$r = 50\ \mathrm{cm}$, $\rho = 10\ \mathrm{g/cm^3}$

Marking scheme:

Points are added additively (including negative points), except for blocks of grey background, where the option with maximal points should be chosen (in absolute value)

A.2: How far one can see from on top of the tower can be related to the radius of the planet via the right triangle shown in the figure below. Applying the Pythagoras theorem on the triangle, one gets $(R+H)^2 = L^2 + R^2$ and so

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A.3: From Newton’s law of gravity, $g = GM/R^2$. Hence,

By adding the errors in quadrature, we find the error

Our estimation of free-fall acceleration has a contribution from the centrifugal force caused by the rotation of the planet. This serves to reduce the acceleration on the surface and hence decrease our estimation of the planet’s mass.

Marking scheme:

The student can’t get overall negative points for value nor error (for example when the value is completely out of range and the units are wrong).

B.1: In general, if the variations in gravitational acceleration are small (as is the case here as $H \ll R$), as a response to air drag, objects tend to terminal velocity where they experience no net acceleration. In the reference frame of air, this corresponds to the object falling straight down with some terminal speed $v_t$. In the lab frame, the object then has horizontal and vertical speeds of $u$ and $v_t$ respectively.

In order to find $u$, we can choose to drop an object that reaches terminal velocity as fast as possible and then observe how the displacement $s$ relates to the fall time $t$. When terminal velocity is reached, we expect $s = s_0 + ut$, where $s_0$ captures the displacement related to reaching terminal velocity. To maximize the effects of air drag, we minimize radius and density, i.e. setting $\rho = 0.1\ \mathrm{g/cm^3}$, and $r = 5\ \mathrm{cm}$. Plotting $s$ vs $t$, we measure the slope to be $u = 1.31\ \mathrm{m/s}$ with an error of $\Delta u = 0.04\ \mathrm{m/s}$.

$r = 5\ \mathrm{cm}$, $\rho = 0.1\ \mathrm{g/cm^3}$

Marking scheme:

B.2: By keeping the measurements close to the surface, we can assume to a good approximation uniform air density. Then, using similar reasoning as before, we expect $h = h_0 + v_{t0}t$, where $h_0$ captures the part of reaching terminal velocity.

At terminal velocity, the drag force balances out gravitational acceleration:

Using $m = 4\pi\rho r^3/3$ and $A = \pi r^2$, we get

On the surface, $v_{t0} = v_t(\rho_a = \rho_{a0})$. Using the measurements from the last subtask, we can plot $t$ vs $h$ and measure the slope to be $1/v_{t0} = 0.037\ \mathrm{s/m}$ with an error of $\Delta(1/v_{t0}) = 0.002\ \mathrm{s/m}$. Hence, $v_{t0} = 27.0\ \mathrm{m/s}$, $\Delta v_{t0} = \Delta(1/v_{t0})/v_{t0}^2 = 2\ \mathrm{m/s}$. Now,

and the error is

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B.3: Due to the adiabatic profile of the atmosphere, the further up you go, the more the temperature and air density decreases, but the terminal velocity increases. We can estimate the terminal velocity of the ball at different heights by comparing the dropping time of a ball with the smallest possible terminal velocity (so minimal density and radius). This hence gives a direct probe for the air density and thus the height of the atmosphere.

If the ball reaches terminal velocity instantly, then the difference in falling time between dropping the ball at heights $h_1$ and $h_2 > h_1$ comes simply from $h_1 < h < h_2$. This is because in both cases the ball falls for the same amount of time at $h < h_1$ (because the terminal velocity only depends on height). Then, if $h_2 - h_1 \ll h_1$, we can estimate

In reality, the ball doesn’t reach the terminal velocity instantaneously. However, it turns out we can, to a good approximation, neglect this effect. As a rough order of magnitude estimation, on the ground level, the ball experiences a time difference of $v_{t0}/(2g) = 0.8\ \mathrm{s}$ compared to the instantaneous case. This difference will increase as the ball is dropped from further up, but as long as the atmosphere isn’t too much sparser in the upper parts of the tower (we can verify this later), the difference will be insignificant compared to the total falling time of the ball. Hence, we approximate the terminal velocity via equation (1).

Because the calculated velocities are very sensitive on the measured quantities, we do repeated measurements throughout the whole height of the tower.

$r = 5\ \mathrm{cm}$, $\rho = 0.1\ \mathrm{g/cm^3}$

Using equation (1) we make a separate table with velocities, while also adding the ground level velocity found in one of the earlier part (we set it at $h = 100\ \mathrm{m}$ because that was the centre of the range of measurements). We find air density using

From the density profile of an adiabatic atmosphere,

Hence, we find $H_0$ by plotting $\rho_{a0}^{0.4}$ against $h$ and fitting a straight line.

$r = 5\ \mathrm{cm}$, $\rho = 0.1\ \mathrm{g/cm^3}$

From the plot, we measure the slope $a = -\rho_{a0}^{0.4}/H_0 = -1.1 \times 10^{-4}\ (\mathrm{kg/m^3})^{0.4}/\mathrm{m}$ and the intercept $b = \rho_{a0}^{2.5} = 0.82\ (\mathrm{kg/m^3})^{0.4}$ so $H_0 = -b/a = 7500\ \mathrm{m}$. We calculate the error from two reasonably chosen lines that correspond to maximal and minimal estimates for $H_0$

We can also confirm that our assumption about the density of the atmosphere not dropping significantly in the upper parts of the tower holds true.

Marking scheme:

B.4: From the expression for adiabatic atmosphere we have

so

and

From ideal gas law,

and

Marking scheme:

C.1: Our goal is to find the rotation speed $\Omega$ of the planet. The rotation of the planet affects the ball’s trajectory via centrifugal and Coriolis force. The centrifugal force, however, due to $H \ll R$ is impossible to disentangle from gravitational acceleration. Coriolis force affects the ball via acceleration $\vec{a}_{\mathrm{cor}} = -2\vec{\Omega} \times \vec{v}$. This is perpendicular to both the velocity of the ball and rotation axis of the planet. Hence, it’s directed along the equator, and increases linearly with the falling speed. Thus, the horizontal acceleration is given by $a_x = 2\Omega v_y + a_{\mathrm{drag}}$.

The procedure is then to minimize the effect of air drag (maximal radius and density) and hope that the Coriolis effect contributes enough to the horizontal displacement. If we neglect air drag, then $a_x = 2\Omega v_y = 2\Omega g t$ so $v_x = \int a_x \mathrm{d}t = \Omega g t^2$ and $x = \int v_x \mathrm{d}t \Omega g t^3/3$. The final displacement will then be $s = \Omega t_f^3/3$, where the falling time satisfies $H = g t_f^2/2$. Putting them together, we get

By varying the radius/density, we do indeed confirm that the effect of Coriolis force is significant, on the order of couple of meters. By doing a suitable number of measurements in the range $0$ to $2000\ \mathrm{m}$ and plotting $s$ vs $h^{1.5}$, we measure the slope

and the error

such that

and

$r = 50\ \mathrm{cm}$, $\rho = 10\ \mathrm{g/cm^3}$

Alternative solution.

An alternative approach is to consider the system in the non-rotating frame (where we don’t have to deal with fictitious forces). In there, the ball starts off with speed $v_0 = \Omega(R+H)$. Due to the conservation of angular momentum, as the ball drops towards the ground, the ball’s angular speed will start increasing and the ground will start lagging behind (the ground rotates with $\Omega$). At height $h$, when the ball moves with angular speed $\omega$, the conservation of angular momentum reads $\omega(R+h)^2 = \Omega(R+H)^2$ and so the angular lag between the ball and the ground is

The positional velocity shift along the ground is then $v_x = \Delta\omega R = 2\Omega(H-h) = \Omega g t^2$. We recover the same expression as for Coriolis force, and from there we proceed the same way as before.

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