Question

A force of 585 N is exerted on a 407 kg mass a distance of 13660 km above the surface of a planet having a mass of 7.9E24 kg. Determine the average density of the planet in kg/cubic meter. Derive and express algebraic solution in terms of givens: F, m, mp, alt and G.

Answer 1

The average density of a planet is given by:

`\rho=(m)/(V)`

where m is the mass of the planet and V is its volume. We know the mass of the planet but we don't know its volume, to find it we will need find its radius.

To find the radius of the planet we can use Newton's Law of gravitation:

`F=G(mM)/(d^2)`

where G is the gravitational constant, m is the mass of the object, M is the mass of the planet and d is the distance between the planet and the object. Let r be the radius of the planet, and x be the distance from the surface of the planet to the object (x=13660 in this case); then we have:

`\begin{gathered} F=G(mM)/((r+x)^2) \\ (r+x)^2=(GmM)/(F) \\ r+x=\pm\sqrt[]{(GmM)/(F)} \\ r=-x\pm\sqrt[]{(GmM)/(F)} \end{gathered}`

Plugging the values given we have:

`\begin{gathered} r=-13660*10^3\pm\sqrt[]{((6.67*10^(-11))(407)(7.9*10^(24)))/(585)} \\ \text{ Using the positive root we have:} \\ r=5.49*10^6 \\ \text{ Using the negative root we have:} \\ r=-3.28*10^7 \end{gathered}`

Since the radius of the planet has to be positive we choose the positive solution.

Now, that we know the radius of the planet we can calculate its volume; assuming the planet is spherical we have that:

`V=(4)/(3)\pi r^3`

then we have:

`\begin{gathered} V=(4)/(3)\pi(5.49*10^6)^3 \\ V=6.92*10^(20) \end{gathered}`

Finally we can calculate the density:

`\begin{gathered} \rho=(7.9*10^(24))/(6.92*10^(20)) \\ \rho=11398 \end{gathered}`

Therefore, the average density is 11398 kg/m^3