Question

question a volume of rock has a mass roughly three times a similar volume of ice. suppose one world is made of ice whereas another world with the same radius is made of rock. if g is the acceleration of gravity on the surface of the ice world, what is the approximate acceleration of gravity on the rock world?

practice it
use the worked example above to help you solve this problem. an astronaut on a space mission lands on a planet with three times the mass and three times the radius of earth. what is her weight wₓ on this planet as a multiple of her earth weight w_e?
wₓ = w_e

exercise
hints: getting started | im stuck!
an astronaut lands on a moon that has one - thirtieth the mass of earth and one - third the radius. find the weight of the astronaut standing on this moon in terms of his earth weight w_e.
wₓ = w_e

resources
read it

Explanation:

Step1: Recall gravitational - force formula

The gravitational force on the surface of a planet is $F = G\frac{Mm}{R^{2}}$, and since $F = mg$, the acceleration due to gravity $g=G\frac{M}{R^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the planet, $m$ is the mass of the object on the planet's surface, and $R$ is the radius of the planet.

Step2: Solve the first problem

Let the mass of the ice - world be $M_{1}$ and the mass of the rock - world be $M_{2}$. Given that for the same volume, $M_{2}=3M_{1}$, and $R_{1} = R_{2}=R$. The acceleration due to gravity on the ice - world $g_{1}=G\frac{M_{1}}{R^{2}}$ and on the rock - world $g_{2}=G\frac{M_{2}}{R^{2}}$. Substituting $M_{2}=3M_{1}$ into the formula for $g_{2}$, we get $g_{2}=3G\frac{M_{1}}{R^{2}} = 3g_{1}$. So the acceleration of gravity on the rock world is $3g$.

Step3: Solve the Practice It problem

On Earth, $w_{E}=mg_{E}=G\frac{M_{E}m}{R_{E}^{2}}$. On the new planet, $M_{X}=3M_{E}$ and $R_{X}=3R_{E}$. The weight on the new planet $w_{X}=mg_{X}=G\frac{M_{X}m}{R_{X}^{2}}$. Substitute $M_{X}=3M_{E}$ and $R_{X}=3R_{E}$ into the formula for $w_{X}$:
\[

$$\begin{align*} w_{X}&=G\frac{3M_{E}m}{(3R_{E})^{2}}\\ &=G\frac{3M_{E}m}{9R_{E}^{2}}\\ &=\frac{1}{3}G\frac{M_{E}m}{R_{E}^{2}} \end{align*}$$

\]
So $w_{X}=\frac{1}{3}w_{E}$.

Step4: Solve the Exercise problem

On Earth, $w_{E}=G\frac{M_{E}m}{R_{E}^{2}}$. On the moon, $M_{X}=\frac{1}{30}M_{E}$ and $R_{X}=\frac{1}{3}R_{E}$. The weight on the moon $w_{X}=G\frac{M_{X}m}{R_{X}^{2}}$. Substitute $M_{X}=\frac{1}{30}M_{E}$ and $R_{X}=\frac{1}{3}R_{E}$ into the formula for $w_{X}$:
\[

$$\begin{align*} w_{X}&=G\frac{\frac{1}{30}M_{E}m}{(\frac{1}{3}R_{E})^{2}}\\ &=G\frac{\frac{1}{30}M_{E}m}{\frac{1}{9}R_{E}^{2}}\\ &=\frac{3}{10}G\frac{M_{E}m}{R_{E}^{2}} \end{align*}$$

\]
So $w_{X}=\frac{3}{10}w_{E}$.

Answer:

For the first question: $3$
For the Practice It problem: $\frac{1}{3}$
For the Exercise problem: $\frac{3}{10}$