0

Usually when on Earth we can use as an approximation:

g = a = 9.18 $\frac{m}{s^2}$

s = vi + 1/2 a$t^{2}$

However, how would I determine the distance traveled in situations where the the acceleration due to gravity is changing non-negligibly because the distances are so large?

I know that the following equation has to be used somehow:

a = $Gmr^{-2}$

I am not sure exactly how though.

For example, if given the information that an object is at a certain distance and is being attracted by a larger object of a given mass, the method should calculate where the object will be at a given time.

Community
  • 1
The Z
  • 101

1 Answers1

1

When a small mass $m$ is moving under the gravitational influence of a large mass $M$, the differential equation that determines the motion $\vec{r}(t)$ of $m$ is

$$\frac{d^2\vec{r}}{dt^2}=-\frac{GM\vec{r}}{r^3},$$

which comes from $\vec{F}=m\vec{a}$ combined with Newton’s Law of Gravitation.

It can be solved, and the solutions for bound orbits are ellipses. (This is of course why planetary orbits are essentially elliptical.) There are also solutions for radial infall, which is just a degenerate ellipse.

You’ve probably solved problems using the often-useful approximation of uniform gravitational acceleration $\vec{g}$ — such as the motion of a cannonball — where the trajectory is a parabola. Those parabolic arcs are all just good approximations to the actual elliptical arcs under Earth’s true inverse-square gravity outside its surface.

G. Smith
  • 52,489