Astronomy Question: Neutron Star Gravitation
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Nicholas writes in with an interesting question regarding Neutron Stars:
The question was a little long, so I’ll summarize his bullet points along with the short answers:
Ahh…. the joys of cosmology and physics. Let’s start with a little math – I won’t go into full calculations, but will show enough to get the basic idea. This will be a long post, so click the “read more” link below to see the full response:
For your first question: Regardless if you are orbiting The Moon, Earth, or a Neutron Star, it’s possible, technically to orbit anything with mass greater than that of ship. Most items in earth orbit are around 17,000 Miles per hour, or roughly 7.5 meters per second.
To calculate orbital speed, more or less you need some information.
Radius of object you are orbiting (Ro) in meters
Mass of object you are orbiting (Mo) in kilograms
Distance from object you are orbiting (D) in meters
The Gravitational Constant, G (6.67×10^-11 N*m2 /kg^2 )
The sum of Ro above and D (R) in meters.
To solve for orbital velocity, we use this formula V = square root of ( G * Mo / R )
If you had a set velocity, you could solve for the radius of the orbit, then subtract the radius of the object to determine how far from the object you would be orbiting.
A Neutron star has anywhere from a little over the mass of the sun to about twice the mass of the sun in a radius of a small city (10-15km) For our example, let’s use a 1.5 Solar Mass Neutron Star ( 3×10^30 kg) with a radius of 15km. (15000m)
Let’s play it safe and orbit this incredibly dense object at the same distance The Moon orbits Earth ( 3.6×10^8 meters )
At this large of a distance, we can just ignore the 15km radius of the Neutron Star.
Note: I did these calculations pretty quickly and they “look” right, but please excuse any minor errors. These equations are basically “napkin math” for lack of a better term.
So, V = square root of [ ( 6.67x10^-11 * 3x10^30 ) / 3.6x10^8 ]
This works out to almost 750,000 meters per second.
In comparison, the fastest object made by man (Helios 2) reached about 70,000 meters per second.
We can further play with this equation, and estimate an orbital radius, given an orbiting speed of .1C:
( 1/10th the speed of light in a vacuum)
3×10^7 = square root of: [ ( 6.67x10^-11 * 3x10^30 ) / 15000+D] gives us R of 207,333 meters.
Subtracting the radius of the Neutron Star (15,000 meters or 15km ) gives us an orbital radius of 192,333 meters (192 km), or roughly 120 miles.
The equation we are using gives velocities for stable orbits, meaning at said velocity, passengers in the rocket ship would feel no different than Astronauts in the Space Shuttle feel when orbiting Earth (effects of being 120 miles from a Neutron Star notwithstanding)
As for your third question, anything in motion experiences time dilation. An experiment using a pair of super-precise atomic clocks proved this many years ago. One atomic clock stayed on the ground, the other flew in a 747. As a matter of fact, GPS satellites have to adjust for this effect in order to keep their clocks synchronized with those on Earth. Mercury’s orbit is also influenced by this phenomenon. So basically, if GPS satellites and Mercury experience effects from their velocities, it would be safe to assume the passengers travelling at 1/10th the speed of light would experience some effects as well.
You can read more about relativistic effects at: http://www.as.utexas.edu/~gebhardt/a309s11/relativity2.html
Hope this helped explain a few things for you!
Ray Sanders is a Sci-Fi geek, astronomer and blogger. Currently researching variable stars at Arizona State University, he writes for Universe Today, The Planetary Society blog, and his own blog, Dear Astronomer
2012-12-04 08:06:04
Source: http://www.dearastronomer.com/2011/05/20/neutron-star-gravitation/
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