Question

Question: I'm trying to figure the acceleration due to gravity of the Earth around the Sun. It orbits at a speed of about 2.58 million km/day. I reason that after half a year, it must be moving an equal speed in the opposite direction. So the change in velocity should be 5.61 mil. km/day over 182.5 days. This works out to about 30740 km/day squared, or about 356 meters/second squared. But I see that this isn't the correct answer. I can't find the flaw in my reasoning, where is it?

Answer 1

The problem is the definition of acceleration. If
a= acceleration
v = velocity and
t = time, then,
\[a = dv/dt = \lim_{\Delta t \rightarrow 0} \Delta v/\Delta t\]
You simply forgot to consider the limiting value of delta t.

To elaborate this, let me extend your own logic. In one year (365 days), the Earth's velocity returns to original value. Thus, change in velocity is zero. So, according to your logic, acceleration must be zero!!!

Hope this helps!

Answer 2

It does help, thank you!

Answer 3

ahh! It should be noted that while the tangential acceleration of the Earth is zero there is most definitely a radial component of acceleration from the Earth to the Sun. This component can be found by from the force due to gravity equation:

\[\left| F_g \right|= G \frac{m_{1} m_{2}}{r}\]

if the mass of the earth is divided out to obtain acceleration, the describing expression is

\[ a_{g, earth} = G \frac{m_{sun}}{r}\]

where G is the gravitational constant, m is the mass of the Sun, and r is the distance from the Sun to the Earth. This acceleration is a vector that points from the Earth to the Sun and is the reason that the Earth revolves around the Sun instead of traveling with constant velocity in a straight line.

Answer 4

You mean radius squared, right?

Answer 5

You're absolutely right! I grabbed an orbital mechanics books off the shelf to make sure I got the equation right but the book apparently had a misprint on the first page! Thanks for the correction.