1. slides 33 - 35 of the powerpoint describes newtons law of universal gravitation: $f_g =…

1. slides 33 - 35 of the powerpoint describes newtons law of universal gravitation: $f_g = g\\frac{m_1m_2}{r^2}$, where $g = 6.67\\times10^{-11} n\\cdot m^2/kg^2$ is the universal gravitational constant. considering that on earth you are subject to the acceleration of gravity $g = 9.81 m/s^2$, and $f_g = mg$, where $m$ is your mass in $kg$, $g = g\\frac{m_e}{r_e^2}$, where $m_e$ and $r_e$ are the mass and radius of the earth, respectively. if you weigh $1/6^{th}$ as much on the moon, calculate the mass of the moon $m_m$ using the radius of the moon $r_m$, and vice versa. check your calculated values against their knowns, how close are they to one another? is it a good estimate to say you weigh $1/6^{th}$ as much on the moon? you can find the values of $m_e, m_m, r_e$, and $r_m$ on this nasa reference page: https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

1. slides 33 - 35 of the powerpoint describes newtons law of universal gravitation: $f_g = g\\frac{m_1m_2}{r^2}$, where $g = 6.67\\times10^{-11} n\\cdot m^2/kg^2$ is the universal gravitational constant. considering that on earth you are subject to the acceleration of gravity $g = 9.81 m/s^2$, and $f_g = mg$, where $m$ is your mass in $kg$, $g = g\\frac{m_e}{r_e^2}$, where $m_e$ and $r_e$ are the mass and radius of the earth, respectively. if you weigh $1/6^{th}$ as much on the moon, calculate the mass of the moon $m_m$ using the radius of the moon $r_m$, and vice versa. check your calculated values against their knowns, how close are they to one another? is it a good estimate to say you weigh $1/6^{th}$ as much on the moon? you can find the values of $m_e, m_m, r_e$, and $r_m$ on this nasa reference page: https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

Answer

Explanation:

Step1: Set up the weight - gravity relationship on Earth and Moon

On Earth, $F_{gE}=mg = G\frac{M_Em}{R_E^{2}}$, so $g = G\frac{M_E}{R_E^{2}}$. On the Moon, $F_{gM}=mg_M=G\frac{M_mm}{R_m^{2}}$, so $g_M = G\frac{M_m}{R_m^{2}}$. We know that $F_{gM}=\frac{1}{6}F_{gE}$, so $g_M=\frac{1}{6}g$.

Step2: Express the mass of the Moon

From $g_M = G\frac{M_m}{R_m^{2}}$ and $g_M=\frac{1}{6}g$, and $g = G\frac{M_E}{R_E^{2}}$, we substitute $g_M$ and $g$: [G\frac{M_m}{R_m^{2}}=\frac{1}{6}G\frac{M_E}{R_E^{2}}] [M_m=\frac{1}{6}\frac{R_m^{2}}{R_E^{2}}M_E] We know from NASA: $M_E = 5.972\times 10^{24}\ kg$, $R_E= 6371\ km = 6.371\times 10^{6}\ m$, $R_m = 1737.4\ km=1.7374\times 10^{6}\ m$. [M_m=\frac{1}{6}\times\left(\frac{1.7374\times 10^{6}}{6.371\times 10^{6}}\right)^{2}\times5.972\times 10^{24}] [M_m=\frac{1}{6}\times\left(\frac{1.7374}{6.371}\right)^{2}\times5.972\times 10^{24}] [M_m=\frac{1}{6}\times0.0746\times5.972\times 10^{24}] [M_m = 0.0124\times5.972\times 10^{24}] [M_m\approx7.41\times 10^{22}\ kg] The known mass of the Moon from NASA is $M_m = 7.348\times 10^{22}\ kg$. The percentage difference $\Delta=\frac{|7.41\times 10^{22}-7.348\times 10^{22}|}{7.348\times 10^{22}}\times 100%=\frac{0.062\times 10^{22}}{7.348\times 10^{22}}\times 100%\approx0.84%$

Answer:

The calculated mass of the Moon is approximately $7.41\times 10^{22}\ kg$, and the known mass from NASA is $7.348\times 10^{22}\ kg$. The percentage - difference between the calculated and known values is about $0.84%$. It is a very good estimate to say you weigh $\frac{1}{6}$th as much on the Moon since the calculated and known values for the mass of the Moon (used to derive the gravitational acceleration on the Moon) are very close.