Question

You are exploring a newly discovered planet. The radius of the planet is 7.50 × 107 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 0.0645 s for a transverse pulse to travel from the lower end to the upper end of the string. On earth, for the same string and lead weight, it takes 0.0320 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that its effect on the tension in the string can be neglected.Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?Express your answer with the appropriate units.

Answer 1

Answer:

$M_{m}=2.31x10^{25} kg$

Explanation:

The distance is the same in the experiment and using the differents times can find the velocities so

$v_{1}=\frac{x}{t_{1}}=\frac{4m}{0.0645s}$

$v_{1}=62.01\frac{m}{s}$

$v_{2}=\frac{x}{t_{2}}=\frac{4m}{0.0320s}$

$v_{1}=125\frac{m}{s}$

Now the experiment can use the equations of tension in simple pendulum

$v=\sqrt{\frac{T*L}{m} }$

$T=\frac{v^2*m}{L}$

$T_{1}=\frac{v_{1}^2*0.0280kg}{4m}$

$T_{1}=\frac{62.01^2*0.0280kg}{4m}$

$T_{1}=26.91N$

$T_{2}=\frac{v_{2}^2*0.0280kg}{4m}$

$T_{12}=\frac{125^2*0.0280kg}{4m}$

$T_{2}=109.375 N$

Now to determinate the mass using the formula of force of gravity

$F_{g}=\frac{G*M_{m}}{r^2}$

$G=6.67x10^{-11} \frac{N*M^2}{kg^2}$

Solve to Mm

$M_{m}=\frac{F_{g}*r^2}{G}$

$M_{m}=\frac{\frac{T_{1}}{T_{2}}*(7.50x10^7m)^2}{6.67x10^{-11}\frac{N*m^2}{kg^2} }$

$M_{m}=\frac{0.273N*5.625x10^15m^2}{6.67x10^{-11}\frac{N*m^2}{kg^2}}$

$M_{m}=2.31x10^{25} kg$