Let's solve the problem step-by-step.
### (i) Ratio of an Object's Weight on the Moon to its Weight on the Earth
The weight of an object is given by [tex]\[W = mg\][/tex], where:
- [tex]\(m\)[/tex] is the mass of the object.
- [tex]\(g\)[/tex] is the acceleration due to gravity.
Given that the gravitational force is proportional to the mass of the planet and inversely proportional to the square of its radius, the weight of an object on the moon (W_moon) and on the earth (W_earth) can be compared.
Given data:
- The mass of the moon is [tex]\(0.0123\)[/tex] times the mass of the earth ([tex]\(m_m = 0.0123 m_e\)[/tex]).
- The radius of the moon ([tex]\(R_{\text{moon}}\)[/tex]) is [tex]\(0.27\)[/tex] times the radius of the earth ([tex]\(R_{\text{moon}} = 0.27 R_{\text{earth}}\)[/tex]).
The weight of the object on the moon:
[tex]\[ W_{\text{moon}} = m \cdot g_{\text{moon}} \][/tex]
The weight of the object on the earth:
[tex]\[ W_{\text{earth}} = m \cdot g_{\text{earth}} \][/tex]
The gravitational acceleration on the moon ([tex]\(g_{\text{moon}}\)[/tex]) and earth ([tex]\(g_{\text{earth}}\)[/tex]) are related by the mass and radius:
[tex]\[ g_{\text{moon}} = g_{\text{earth}} \cdot \left(\frac{m_m}{m_e} \cdot \frac{R_{\text{earth}}^2}{R_{\text{moon}}^2}\right) = g_{\text{earth}} \cdot \left(\frac{0.0123}{1} \cdot \frac{1}{(0.27)^2}\right) \][/tex]
Calculating the above ratio:
[tex]\[ \frac{W_{\text{moon}}}{W_{\text{earth}}} = \frac{g_{\text{moon}}}{g_{\text{earth}}} \][/tex]
[tex]\[ = 0.0123 \cdot \frac{1}{(0.27)^2} \][/tex]
[tex]\[ \approx 0.00089667 \][/tex]
So, the ratio of the object's weight on the moon to its weight on the earth is approximately [tex]\(0.00089667\)[/tex].
### (ii) Acceleration of Free Fall on the Moon
We need to find the acceleration of free fall on the moon ([tex]\(g_{\text{moon}}\)[/tex]).
We already derived the equation for [tex]\(g_{\text{moon}}\)[/tex]:
[tex]\[ g_{\text{moon}} = g_{\text{earth}} \cdot \left(\frac{R_{\text{moon}}}{R_{\text{earth}}}\right)^2 \][/tex]
Given:
[tex]\[ g_{\text{earth}} = 9.8 \, \text{m/s}^2 \][/tex]
Substitute the given values:
[tex]\[ g_{\text{moon}} = 9.8 \, \text{m/s}^2 \cdot (0.27)^2 \][/tex]
[tex]\[ g_{\text{moon}} \approx 0.71442 \, \text{m/s}^2 \][/tex]
Therefore, the acceleration of free fall on the moon is approximately [tex]\(0.71442 \, \text{m/s}^2\)[/tex].
