Question

Because your spaceship has an engine failure, you crash-land with an emergency capsule at the equator of a nearby planet. The planet is very small and the surface is a desert with some stones and small rocks laying around. You need water to survive. However, water is only available at the poles of the planet. You find the following items in your emergency capsule: • Stopwatch • Electronic Scale • 2 yardstick • 1 Litre Oil • Measuring Cup Describe an experiment to determine your distance to the poles by using the available items. Hint: As the planet is very small, you can assume the same density everywhere.

Answer 1

Answer:

The distance from the poled, $D_p$, is given as follows;

$D_p = \dfrac{ 4 \cdot M \cdot L \cdot \pi ^2 - P \cdot T^2\cdot 3}{24 \cdot M \cdot \pi }$

Explanation:

The materials available in the emergency capsule are;

1) Stopwatch

2) Electronic scale

3) 2 yardstick

4) 1 Litre Oil

5) Measuring Cup

We note that the distance from the poles from the equator = $\dfrac{1}{4} \times 2 \pi R$

We place some known quantity of oil in the cup and weigh with the electronic scale

Where the weight is P Newtons we have at the equator;

P = Weight of cup + oil - centripetal force on cup + oil

$P = M \times g_p - \dfrac{M \times v^2}{R} = M \times g_p - \dfrac{4 \times M \times \pi ^2 \times R}{T^2}$

Where:

M = The mass of the cup + the mass of the Oil

$g_p$ = Acceleration due to gravity on the planet

R = Radius of the planet

T = Period of rotation of the planet

The mass of the oil can be found by measuring the depth to which the cup sinks when placed in the oil

The depth of immersion × cross-sectional area of the cup  found with the yardstick gives the volume, $V_{cup}$, of the oil equivalent to the mass of the cup from which the density of the oil is calculated as follows;

$\rho_{oil} = \dfrac{m_{cup}}{V_{cup}}$

The mass of the oil = $V_{oil} \times \rho_{oil}$

The mass of the cup can be be found on the tag or engraved on the cup

Which gives the radius of the planet as follows;

$R = \dfrac{(M \times g_p - P) }{ M \times v ^2}$

or

$R = \dfrac{(M \times g_p - P) \times T^2}{4 \times M \times \pi ^2}$

The period or the time it takes the planet to make one revolution can be found by measurement with the stopwatch

The acceleration due to gravity on the planet is found from the relation of the simple pendulum formed by the yard stick as follows;

$T = 2 \pi \sqrt{\dfrac{I}{m_y \times g_p \times L} }$

Where:

$m_y$ = Mass of the yardstick

L = Yardstick length

I = Moment of inertia of the yardstick

$I = \dfrac{m_y \times L^2}{3}$

Which gives;

$T = 2 \pi \sqrt{\dfrac{L}{3 \times g_p} }$

From which;

$g_p = \dfrac{4 \times \pi ^2 \times L}{3 \times T^2}$

Substituting in the equation for R, we have;

$R = \dfrac{ 4 \times M \times L \times \pi ^2 - P \times T^2\times 3}{12 \times M \times \pi ^2}$

The distance, $D_p$, from the poles = $\dfrac{1}{4} \times 2 \pi R$= $\dfrac{ \pi}{2} \times R$

Therefore;

$D_p = \dfrac{ 4 \times M \times L \times \pi ^2 - P \times T^2\times 3}{12 \times M \times \pi ^2} \times \dfrac{\pi }{2} = \dfrac{ 4 \times M \times L \times \pi ^2 - P \times T^2\times 3}{24 \times M \times \pi }$