Question

A future space explorer has been kidnapped and is held prisoner on a planet in our solar system. With nothing else to do, our prisoner amuses herself by dropping her watch from eye level (170 cm) to the floor. She observes that the watch takes 0.36 s to fall. Ignore air resistance.On which of the following planets is she most likely being held? a. Saturn b. Pluto c. Venus d. Jupiter

Answer 1

To develop this problem we will apply the linear motion kinematic equations. Specifically, the second law that describes the position of a body as a function of its initial velocity, time and acceleration.

$y = ut+\frac{1}{2}gt^2$

Here,

u = Initial velocity

t = Time

g = Acceleration due to gravitation

If we replace the values to find the gravitational acceleration we have then,

$1.7m = 0+\frac{1}{2} g*0.36^2$

$g= 26.2346m/s^2$

Recall that the force of gravity on the planet Jupiter is 24.79 m / s² so the measure is closer to this planet. It is likely that you are in Jupiter.

Answer 2

The space explorer is held prisoner on the planet Jupiter as the acceleration due to gravity calculated matches with that of Jupiter.

Here, the watch is dropped from 170 m above the floor and it undergoes free fall. The watch reaches the floor in 0.36 s.

• Initial velocity $u= 0\,m/s$
• Displacement of the watch $s = 170\, cm = 1.7\,m$
• Time taken $t=0.36\,s$

As the motion is free fall, we can use the kinematics equation by replacing 'a' with '$g_p$'. Where '$g_p$' is the acceleration due to gravity of that planet.

• $s=ut+\frac{1}{2}\,g_p t^2$

• But $u= 0\,m/s$

• Therefore, $s = \frac{1}{2} \,g_p t^2$

• ie; $g_p = \frac{2s}{t^2} = \frac{2\times 1.70\,m}{(0.36\,s)^2} = 26.23\, m/s^2$

The acceleration due to gravity of Jupiter is found to be $24.5\,m/s^2$.

Therefore, the value of acceleration due to gravity we have calculated is closest to acceleration due to gravity of Jupiter.

Hence we can say that the space explorer is held on Jupiter.

Learn more about free fall here:

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