What is the radial acceleration, in , of the earth toward the sun?

Question

What is the magnitude of the orbital velocity, in , of the earth around the sun?What is the radial acceleration, in , of the earth toward the sun?

Answer 1

There are a couple of ways one could approach this problem all leading to the same answer. We will assume the Earth orbits in a circle around the Sun instead of a slight elipse.

The obvious one, the method you are probably thinking of, involved finding the orbital speed of the Earth around the sun (v) and using the radial distance of the Earth from the Sun, find its centripetal acceleration.
a_c = v^2 / r
To use this method one needs to know the value of v and r though. Not to worry, NASA will help us with those.
v = [average] orbital velocity of the Earth around the sun = 29.8 km/s = 29800 m/s
r = [average] orbital distance of Earth from Sun = 149.6 E6 km = 149.6 E9 meters
So, plugging into the above equation,
a_c = (29800 m/s)^2 / (149.6 E9 meters)
a_c = .00593 m/s^2

Another method (which is essentially the same thing) involves using the angular velocity of Earth around the Sun. This way, all we need to know is the distance of the Earth from the sun and the time it takes the Earth to go around one complete cycle (1 year).

Another method can be used when one thinks of where this centripetal acceleration is coming from. What is providing it? Gravity.
The centripetal acceleration of the Earth around the Sun is equal to the gravitational acceleration the Earth experiences due to the Sun’s mass.
F = G * M / r^2
Where G is the universal gravitational constant, M is the mass of the Sun, and r is the distance from the Earth to the Sun.
G = 6.67 E-11 N m^2 / kg^2
M = 1.99 E30 kg
r = 149.6 E9 meters
Plugging in, we get,
F = (6.67 E-11 N m^2 / kg^2) (1.99 E30 kg) / (149.6 E9 meters)^2
F = .00593 m/s^2

Both (all 3) methods give us the same answer, a = .00593 m/s^2 inward towards the Sun.

Answer 2

The orbital velocity of the earth is roughly
V = (2π*9.3*10^7 mi/365.259 da)(1 da/24 hr)(1 hr/3600 sec) ≈ 18.516 mi/sec.
The acceleration towards the sun is about
a ≈ (18.516^2)/(9.3*10^7) ≈ 3.6865e-6 mi/sec^2