Question

The region of the solar system between Mars and Jupiter contains many asteroids that orbit the Sun. Consider an asteroid in a circular orbit of radius 4.0 1011 m. Find the period of the orbit

Answer 1

4.0e11

Answer 2

The question is presented as follows: No further info is given

The region of the solar system between Mars and Jupiter contains many asteroids that orbit the Sun. Consider an asteroid in a circular orbit of radius 4.0e11 m. Find the period of the orbit

Answer 3

@roadjester any ideas?

Answer 4

Yeah..I'm working on it...

Answer 5

ok

Answer 6

use this equation which assumes the mass orbiting is negligible compared to the sun..which it is:
Vo = sq rt [ GM /r]
G = gravitational acceleration of sun = 274 meters/second
M = mass of sun = 1.9891×10^30 kg
Vo = velocity of orbiting asteroid
r = radius of orbit = 4.0 * 10^11 meters

Velocity of orbit = sq rt [(( 274 )(1.9891*10^30)) / (4*10^11)]
Velocity of asteroid = 3.69125 * 10^10 m/s

Now you need the circumference of that orbit and then the period is the time it takes for that asteroid to circle the sun once:

Circum = 2* pi * r
Circum = 2 *3.14 * 4*10^11
circumference of asteroid orbit = 2.51327 * 10^12 meters

And last, divide distance of orbit by velocity of orbit to get period:
(2.51327 * 10^12 m) / (3.69125 * 10^10 m/s)

period = 68 seconds

That is one fast asteroid

Answer 7

gravitational acceleration of sun = 274 m/s^2....sorry

Answer 8

According to Kepler's Third Law, the orbital period T\, (in seconds) of two bodies orbiting each other in a circular or elliptic orbit is:
T = 2\pi\sqrt{a^3/\mu}
where:
a\, is the orbit's semi-major axis, in kilometers
\mu = GM \, is the standard gravitational parameter, typically in km^3/s^2
G \, is the gravitational constant,
M \, is the mass of the more massive body.

For all ellipses with a given semi-major axis the orbital period is the same, regardless of eccentricity.

Answer 9

http://en.wikipedia.org/wiki/Orbital_period