1
$\begingroup$

Suppose we lived on the surface of a spherical planet, with uniform density. How would the gravitational force we experience change if we drilled a hole and descended towards the centre of the planet?

The magnitude of gravitational attraction between two objects is G=m1*m2/d^2. When considering any two points, this force is 0 (as a point has no mass). For one point in relation to the infinitude of other points in the sphere, there would be a gravitational force. The question is this: for a point with a distance x from the centre of the sphere, what is the sum of the gravitational attraction to all other points in the sphere - that is, what is the net force experienced.

Clearly, at the centre of the sphere (x=0) there is no net force experienced; while at the surface there is a force experienced (towards the centre of the sphere)

$\endgroup$
1
  • 2
    $\begingroup$ There is an analogous question on the electric field inside a uniformly-charged ball. The simplest approach in that case is to use Gauss's law for electrostatics. If you can deduce the gravitational analogue, that can also be used here. $\endgroup$ Jun 23, 2016 at 11:52

1 Answer 1

3
$\begingroup$

As noted in a comment, the most straightforward approach is through Gauss's law. The force on a body outside a spherical shell is the same as if the shell were concentrated at the centre, and the force on a body inside a spherical shell vanishes. Thus, at a distance $x$ from the centre of a sphere of radius $r\ge x$ and mass $m_1$, a body of mass $m_2$ experiences a force

$$ G\frac{\left(\frac{x^3}{r^3}m_1\right)m_2}{x^2}=Gm_1m_2\frac x{r^3}\;, $$

that is, the force increases linearly with the distance from the centre.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .