Q56P - University Physics with Modern Physics

Question

You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times {10^7}\;{\rm{m}}\). You suspend a lead weight from the lower end of a light string that is 4.00 m long and has mass 0.0280 kg. You measure that it takes 0.0685 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that you ignore its effect on the tension in the string. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?

Step-by-Step Solution

Answer

Thus, the mass of the planet is \(29.53 \times {10^{25}}\;kg\).

Step 1: Given in the question

Radius of planet \(r = 7.20 \times {10^7}\;{\rm{m}}\).

 

The mass of wire \(m = 0.0280\;{\rm{kg}}\).

 

Length of string \(L = 4.00\;m\)

 

Time taken for transverse pulse to travel the length of the string \(t = 0.0685\;{\rm{s}}\).

Step 2: Use formula of gravitational acceleration

The velocity of a transverse wave is proportional to root of tension in the string is:

 

\(\begin{array}{l}V \propto \sqrt T \lambda \\V = k\sqrt T \\g = G\frac{M}{{{r^2}}}\end{array}\).

 

Now, write as follows:

 

\(\frac{{{V_1}}}{{{V_2}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \)

 

Here, \({V_1}\) and \({V_2}\) are velocities of wave corresponding to tension of \({T_1}\) and \({T_2}\) in the string.

Step 3: Calculate the mass of planet

Let \({T_1}\) be tension when the experiment was conducted on new planet and \({T_2}\) is tension when it was conducted on the Earth.

 

Tension is equal to the weight suspended so, solve as follows:

\(\frac{{{V_1}}}{{{V_2}}} = \sqrt {\frac{{0.028\;g}}{{0.028 \times 9.8}}} \)

 

Where \(g\) is gravitational acceleration on new planet.

 

\(\begin{array}{c}{V_1} = \frac{4}{{0.0625}}\\ = 64\;{\rm{m/s}}\end{array}\)

 

And

 

\(\begin{array}{c}{V_2} = \frac{4}{{0.039}}\\ = 102.564\;{\rm{m/s}}\end{array}\)

 

Now, solve by putting above values gives:

 

\(\begin{array}{c}\frac{{64}}{{102.564}} = \sqrt {\frac{{0.028\;g}}{{0.028 \times 9.8}}} \\0.389 = \frac{g}{{9.8}}\\g = 3.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

 

Now, use the formula for gravitational acceleration as follows:

 

\(\begin{array}{c}g = \frac{{GM}}{{{r^2}}}\\3.8 = 6.67 \times {10^{ - 11}} \times \frac{M}{{{{\left( {7.2 \times {{10}^7}} \right)}^2}}}\\M = 29.53 \times {10^{25}}\;kg\end{array}\)

 

Here, \(G\) is universal gravitational constant.

 

Hence, the mass of the planet is \(29.53 \times {10^{25}}\;kg\).