91影视

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 \(t = 0.0685\;{\rm{s}}\).

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.

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 \(29.53 \times {10^{25}}\;kg\).