Question

problem 3
the gravitational acceleration of an object on the surface of the moon is approximately 1/6 the gravitational acceleration on the surface of earth. it is also found that the diameter of the moon is d = 3,476 km. what is the mass (m) of the moon? make sure to do the follow the steps 1) draw the free body diagram 2) write down the expression for the force and equate it to mass time acceleration 3) solve for the acceleration 4) write a final expression for mass of the moon in terms of the knowns 5) get a number.

Explanation:

Step1: Draw Free - Body Diagram

An object on the Moon's surface has two forces acting on it: the gravitational force \(F = \frac{GMm}{R^{2}}\) (where \(G\) is the gravitational constant, \(M\) is the mass of the Moon, \(m\) is the mass of the object, and \(R\) is the radius of the Moon) and the normal force \(N\). In the case of an object at rest on the surface (or in free - fall near the surface), \(N = 0\) for the free - fall situation, and the net force is the gravitational force.

Step2: Force and Newton's Second Law

The gravitational force \(F=\frac{GMm}{R^{2}}\), and by Newton's second law \(F = ma\) (where \(a\) is the gravitational acceleration). So, \(\frac{GMm}{R^{2}}=ma\). The mass \(m\) of the object cancels out (since \(m
eq0\)), and we get \(a=\frac{GM}{R^{2}}\).

Step3: Known Values

We know that \(a = \frac{1}{6}g_{E}\) (where \(g_{E}=9.8\ m/s^{2}\)), and \(R=\frac{D}{2}\). Given \(D = 3476\ km=3.476\times 10^{6}\ m\), so \(R = 1.738\times 10^{6}\ m\)

Step4: Solve for \(M\)

From \(a=\frac{GM}{R^{2}}\), we can express \(M=\frac{aR^{2}}{G}\). Substituting \(a=\frac{1}{6}g_{E}\), we get \(M=\frac{g_{E}R^{2}}{6G}\)

Step5: Calculate the Mass

We know that \(G = 6.67\times 10^{- 11}\ N\cdot m^{2}/kg^{2}\), \(g_{E}=9.8\ m/s^{2}\), and \(R = 1.738\times 10^{6}\ m\)

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Answer:

The mass of the Moon is approximately \(M = 7.4\times 10^{22}\ kg\)