Answer:
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 tension of the string between m1 and m3, and 
the tension of the string between m2 and m3, the problem can be solved by writing the following system of equations:
