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

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)