You are exploring a newly discovered planet. The radius of the planet is 7.20 × 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 0.0685 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 0.0370 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.

Part A

Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?
Express your answer with the appropriate units.

Mass of other planet = 2.64 x 10^(26) m

Explanation:

Radius of other planet (R) = 7.2 x 10^(7)m

Mass of string; M= 0.028kg

Length of string, L= 4m

Time on other planet(Tp) = 0.0685 s

Time on earth (Te) = 0.0370 s

First of all, let's find the lead on the earth;

Linear mass density is given by;

μ = M/L = 0.028/4 = 0.007 Kg/m

The speed of the wave here is given by; Ve = L/t = 4/0.037 = 108.11 m/s

Tension in the spring(Fe) is given by the formula ;

Fe = μ(Ve)² = 0.007 x 108.11² = 81.81N

If we apply Newton's second law of motion to this earth lead, we'll arrive at;

ΣFy = Fe - Wl = 0

And so Fe - W(l) = 0 and Fe = W(l)

We know that weight(W) = Mg

Thus; Fe = M(l)g

Where M(l) is mass of the lead; and g is acceleration due to gravity on earth which is 9.81

Thus; M(l) = Fe/g = 81.81/9.81 = 8.34kg

Following the same pattern, let's calculate the lead on the other planet;

The linear density is the property of a material and it remains same as;

μ = 0.007 Kg/m

The speed of the wave here is given by; Vp = L/t = 4/0.0685 = 58. 39 m/s

Tension in the spring(Fp) is given by the formula ;

Fp = μ(Vp)² = 0.007 x 58.39² = 23.87 N

If we apply Newton's second law of motion to this earth lead, we'll arrive at;

ΣFy = Fp - Wl = 0

And so Fe - W(l) = 0 and Fp = W(l)

We know that weight(W) = Mg(p)

Thus; Fp = M(l)g(p)

Where M(l) is mass of the lead; and g(p) is acceleration due to gravity om this other planet

Thus; gp = Fp/M(l) = 28.37/8.34 = 3.4 m/s²

From gravity equation, we know that; acceleration due to gravity of planet is; g = (GM)/r²

Making M the subject, we have;

(gr²)/G = M

Where G is gravitational constant which has a value of 6.6742 x 10^(-11) Nm²/kg²

M is planet mass

r is planet radius

Thus;

M = [3.4 x (7.2 x 10^(7))²]/ 6.6742 x 10^(-11) = 2.64 x 10^(26)m

the answer would be a

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000