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Chapter 2: Problem 68

The planet Mercury has a mass that is \(5 \%\) of that of Earth, and its gravitational acceleration is \(g_{\text {mercury }}=3.7 \mathrm{~m} / \mathrm{s}^{2}\) a) How long does it take for a rock that is dropped from a height of \(1.75 \mathrm{~m}\) to hit the ground on Mercury? b) How does this time compare to the time it takes the same rock to reach the ground on Earth, if dropped from the same height? c) From what height would you have to drop the rock on Earth so that the fall- time on both planets is the same?

Short Answer

Expert verified
Based on the above solution, if a rock falls from a height of 1.75 meters on Mercury, it takes approximately 0.974 seconds for it to hit the ground, while on Earth, it takes approximately 0.598 seconds to fall from the same height. In order to have the same fall-time on both Mercury and Earth, the rock needs to be dropped from a height of approximately 4.625 meters on Earth.

Step by step solution

01

Determine the kinematic equation for the fall-time

We will use the standard kinematic equation for free-fall motion to calculate the fall time. The equation is: \(h = \frac{1}{2}gt^2\). Here, \(h\) is the height, \(g\) is the gravitational acceleration, and \(t\) is the time.
02

Plug in the values for Mercury

We will plug in the values for Mercury: \(h = 1.75~m\) and \(g = 3.7~m/s^2\). By rearranging the equation we get: \(t = \sqrt{\frac{2h}{g}}\).
03

Compute the fall-time on Mercury

Now, we calculate the fall-time for the rock on Mercury: \(t = \sqrt{\frac{2(1.75)}{3.7}} \approx 0.974~s\). b) Time comparison with Earth:
04

Determine Earth's gravitational acceleration

The gravitational acceleration on Earth is: \(g_{\text {earth }}=9.81~m /s^{2}\).
05

Compute the fall-time on Earth

Using the equation from Step 2, we plug in the values for Earth: \(h = 1.75~m\) and \(g_{\text {earth }} = 9.81~m/s^{2}\). Then, we calculate the fall-time for the rock on Earth: \(t_{\text {earth }}=\sqrt{\frac{2(1.75)}{9.81}}\approx 0.598~s\).
06

Compare the fall-times on both planets

It takes approximately 0.974 seconds for the rock to hit the ground on Mercury and 0.598 seconds on Earth. The rock takes longer to fall on Mercury due to its weaker gravitational force. c) Same fall-time on both planets:
07

Set up an equation with equal fall-times

Using the equation from step 2 for both planets, we have: \(\sqrt{\frac{2h_{\text {earth }}}{g_{\text {earth }}}}=\sqrt{\frac{2h_{\text {mercury }}}{g_{\text {mercury }}}}\) Since the fall-time should be the same on both planets, we can equate them: \(\frac{2h_{\text {earth }}}{g_{\text {earth }}}=\frac{2h_{\text {mercury }}}{g_{\text {mercury }}}\)
08

Solve for the height on Earth

Rearrange the equation to find the height on Earth: \(h_{\text {earth }}=\frac{g_{\text {earth }}}{g_{\text {mercury }}}h_{\text {mercury}}\)
09

Plug in the values for Earth and Mercury

We plug in the values for both planets and calculate the height on Earth: \(h_{\text {earth }}=\frac{9.81}{3.7}(1.75) \approx 4.625~m\)
10

Conclusion

The rock has to be dropped from a height of approximately 4.625 meters on Earth in order for the fall-time to be the same as on Mercury.

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