Answer:
Explanation:
If the acceleration due to gravity on the Moon's surface is [tex]a_M[/tex] and the acceleration due to gravity on the Earth is [tex]a_E[/tex], we can write that:
[tex]a_M=\frac{1}{6}a_E[/tex]
[tex]\frac{a_M}{a_E}=\frac{1}{6}[/tex]
If the radius of the Moon is [tex]r_M[/tex] and the radius of the Earth is [tex]r_E[/tex], we can write that:
[tex]r_M=\frac{1}{4}r_E[/tex]
[tex]\frac{r_M}{r_E}=\frac{1}{4}[/tex]
By Newton's 2nd Law we know that F=ma and using Newton's law of universal gravitation we can calculate the gravitational force an object with mass m experiments from a planet with mass M being at a distance r from it. We will assume our object is on the surface so this distance will be the radius of the planet.
Since the force the object experiments is the force of gravitation we can write, for Earth:
[tex]F=ma_E=\frac{GM_Em}{r_E^2}[/tex]
which means:
[tex]a_E=\frac{GM_E}{r_E^2}[/tex]
[tex]M_E=\frac{a_Er_E^2}{G}[/tex]
And for the Moon:
[tex]F=ma_M=\frac{GM_Mm}{r_M^2}[/tex]
which means:
[tex]a_M=\frac{GM_M}{r_M^2}[/tex]
[tex]M_M=\frac{a_Mr_M^2}{G}[/tex]
We can then write the fraction:
[tex]\frac{M_M}{M_E}=\frac{a_Mr_M^2}{G}\frac{G}{a_Er_E^2}=\frac{a_M}{a_E}(\frac{r_M}{r_E})^2=\frac{1}{6}(\frac{1}{4})^2=0.01[/tex]
Which means:
[tex]M_M=0.01M_E[/tex]
