Astronomy C

Test your knowledge of various Science Olympiad events.
astronomybuff
Member
Member
Posts: 36
Joined: January 25th, 2020, 9:19 am
Division: C
State: NC
Has thanked: 1 time
Been thanked: 2 times

Re: Astronomy C

Post by astronomybuff » December 9th, 2020, 9:16 am

0sm0sis wrote:
December 9th, 2020, 8:59 am
Yay orbits!
Because the orbit is circular, we can equate the centripetal acceleration to the gravitational force to find that v^2 = GM/R.

From there, we can use the Vis-Viva equation to find the semimajor axis of the new orbit. The equation is V^2 = GM ( 2/R - 1/a ), but we must use v/2 in place of V. This gives us

GM/4R = 2GM/R - GM/a

Which simplifies to:

4/7 R = a

From there we should use reasoning with how orbits work. Because the velocity is perfectly tangential from the beginning, the initial distance from the star is either the perigee or apogee. Because a larger velocity is required to have circular motion for this position, this distance is probably the apogee, the furthest distance from the star.

We also know that the perigee distance + apogee distance = 2a (this can be seen easily by drawing a diagram of an elliptical orbit). Therefore, R + x = 2(4/7 R), meaning that our perigee x is:

x = R/7
Absolutely correct!. I love orbits too!

I used a slightly different method with energy. The total energy is -GMm/2a. Kinetic energy can be derived using your method of equating centripretal force to gravitational force: mv^2/r=GMm/r^2, mv^2=GMm/r, but we need a 1/2 for kinetic energy, so kinetic energy= GMm/2r. But we have v/2, which makes the kinetic energy GMm/8r. The total energy is kinetc+potential, or GMm/8r-GMm/r= -7/8 GMm/r. Setting this equal to -GMm/2a, we find a= 4/7 R. Then we can use the method you gave us. The total major axis is 8/7 R, and since the satellite started a distance R, we subtract R to get R/7.
These users thanked the author astronomybuff for the post:
0sm0sis (December 9th, 2020, 11:51 am)

User avatar
0sm0sis
Member
Member
Posts: 5
Joined: July 25th, 2020, 3:30 pm
Division: C
State: OH
Has thanked: 2 times
Been thanked: 4 times

Re: Astronomy C

Post by 0sm0sis » December 9th, 2020, 9:53 am

(sorry to people who don't like orbits)

Imagine that the solar system consists of just the Earth circularly orbiting the Sun. If you wanted to get a spaceship to escape the solar system from the surface of Earth, what is the minimum velocity with respect to the Earth needed? Use the values of:

Mass of Earth = 5.97 * 10^24 kg
Radius of Earth = 6400 km
Mass of Sun = 1.99 * 10^30 kg
Distance from Sun to Earth = 1.5 * 10^8 km

(question credit: Morin)
2020-21 Events:

Circuit Lab, Detector, Machines, SOM

"All science is either physics or stamp collecting." - Ernest Rutherford

Post Reply

Return to “Question Marathons”

Who is online

Users browsing this forum: No registered users and 2 guests