Question 331: Gravitation

Question

The average density of the earth is [g is acceleration due to gravity]

Accepted Answer

To determine the relationship between the average density of the Earth and the acceleration due to gravity (g), we start with the formula for gravitational force:
$ F = G \frac{m_1 m_2}{r^2} $
Here, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of two objects, and $r$ is the distance between their centers of mass. For an object near the Earth's surface, this force is also equal to the weight of the object, which is given by:
$ F = m g $
Setting these two expressions for the force equal to each other, we get:
$ m g = G \frac{m M_{	ext{earth}}}{r^2} $
Here, $M_{	ext{earth}}$ is the mass of the Earth, and $r$ (or $R$) is the radius of the Earth. Simplifying, we obtain the equation:
$ g = G \frac{M_{	ext{earth}}}{R^2} $
Now, the mass of the Earth can be expressed in terms of its volume and average density ( $ho_{	ext{earth}}$ ) as follows:
$ M_{	ext{earth}} = ho_{	ext{earth}} \cdot V_{	ext{earth}} $
Since the volume of a sphere is given by
$ V_{	ext{earth}} = \frac{4}{3} \pi R^3 $
we can substitute this into our expression for $M_{	ext{earth}}$ to get:
$ M_{	ext{earth}} = ho_{	ext{earth}} \cdot \frac{4}{3} \pi R^3 $
Substituting this back into our equation for $g$, we have:
$ g = G \frac{ho_{	ext{earth}} \cdot \frac{4}{3} \pi R^3}{R^2} $
Simplifying the equation, we get:
$ g = \frac{4 \pi G}{3} ho_{	ext{earth}} R $
From this equation, we can see that the average density of the Earth, $ho_{	ext{earth}}$, is directly proportional to $g$, i.e.,
$ ho_{	ext{earth}} \propto g $
Hence, the correct answer is:
Option B
directly proportional to $\mathrm{g}$.

Answer

inversely proportional to g$^2$.

Answer

directly proportional to $\mathrm{g}$.

Answer

inversely proportional to $\mathrm{g}$.

Answer

directly proportional to $\mathrm{g}^2$.

Gravitation

The average density of the earth is [g is acceleration due to gravity]

Choose the correct answer: