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2004-May-16, 05:24 PM
What is the formula used to calculate an object's mass in the universe?

Tobin Dax
2004-May-16, 06:23 PM
Could you be more specific or restate the question, please? I'm not exactly sure what you mean.

2004-May-16, 06:32 PM
I think it's based on Kepler's third law of planetary motion. I remember asking this before, but I didn't really understand.

Kullat Nunu
2004-May-16, 06:39 PM
What is the formula used to calculate an object's mass in the universe?

Depends on the object.

Binary objects (binary stars, star-planet, planet-moon, etc.) are easier:

T^2 = 4*Pi^2*d^3/(G*(m1+m2))

where

T = orbital period [s]
d = distance between the two bodies [m]
G = gravitational constant = 6.67 * 10^-11 N*m^2/kg^2
m1, m2 = masses of the two bodies [kg]

Just solve m1 or m2 (you need to know other).

Single objects like stars is much harder cases. We must find out some of its physical properties, like its distance, magnitude and age to figure out its luminosity and theoretical mass. Error bars are naturally large.

If you land on an alien planet whose radius you know you can easily determine its mass just by weighting yourself:

w = G*m_p*m/r^2

w = your weight [kg]
G = same as above
m_p = mass of the planet [kg]
m = your mass [kg]
r = radius of the planet [m]

You can solve an asteroid's mass by determining its composition and volume. They are easier than planets or spherical moons since they are too small to compress gravitationally.

m = rho*V

where

m = mass of the asteroid [kg]
rho = density [kg/m^3] (e.g. asteroid Eros have a density of 2400 kg/m^3)
V = volume [m^3]

If you manage to determine the radius of a black hole, you can derive its mass directly:

R_s = 2*G*M/c^2

where

M = mass of the black hole [kg]
c = 299,792,458 m/s = the speed of light

I think this is enough...

Arrgh... Looks like me too is becoming a formula addict of sorts #-o ...

2004-May-16, 06:41 PM
Wow, did you major in physics or something? :o

harlequin
2004-May-16, 06:45 PM
What is the formula used to calculate an object's mass in the universe?

If I had not seen that you are poster of many previous posts, I would have sworn that you were asking us to do your homework....

In a Newtonian universe:

f=G m1 m2 / r^2

where f is the force of gravity, G is a constant which represents the strength of gravity, m1 and m2 are masses of the two objects attracting each other, and r is the distance between their center of gravities.

G can be determined by experimentation.

One can get the mass of the Earth by

f=ma

where f is force, m is the mass, and is accelleration. At sea level the accelleration due to gravity is about 9.8 meters per second squared and is usually represented by a g. So..

f=mg

Combine with the first equation and using me for mass of the Earth and
mo for mass of falling object:

G mo me / r^2 = mo g

mo cancells out:

G me / r^2 = g

Rearrange and one gets

me = g r^2 / G

We obviously know the distance between the object and the center
of the Earth and that is r.

Since mass of a planet (or whatever) determines how fast an object in x orbit will travel, one can determine masses for other objects as well though the math is a bit more complicated.

There are other things as well. For main sequence stars there is a relationship between how bright the star is and how massive it is.

Kullat Nunu
2004-May-16, 06:49 PM
T^2 = 4*Pi^2*d^3/(G*(m1+m2))

This can be reduced to much simpler form:

M = a^3 / P^2

if

M = mass of the star (much larger than the mass of the planet) in Solar masses
a = distance in AUs and
P = in Earth years

*sigh* almost destroyed my original post!

2004-May-16, 07:06 PM
Binary objects (binary stars, star-planet, planet-moon, etc.) are easier:

T^2 = 4*Pi^2*d^3/(G*(m1+m2))

where

T = orbital period [s]
d = distance between the two bodies [m]
G = gravitational constant = 6.67 * 10^-11 N*m^2/kg^2
m1, m2 = masses of the two bodies [kg]

Just solve m1 or m2 (you need to know other).

Ok, first, I solved for M (I discounted m2 since I am trying to find the mass of the sun, and earth's mass is negligible). The formula is

M=(4*pi^2)d^3/G*P^2, or

M=39.5*d^3/G*P^2.

I wanted to find the mass of the sun, so

M= (39.5)(1.5×10¹¹ m)³
_________________________________
(6.67×10-¹¹ N×m²/kg²)(3.1536×10^7 s)

M=1.33×10^35 m³
_________________
2.1035×10-³ N×m²×s/kg²

M=6.3×10^37 m/N*s/kg^2.

The real mass of the sun is 2.0×10³º kg. In addition, my units did not cancel. Can someone tell me where I went wrong?

Thanks.

Kullat Nunu
2004-May-16, 07:22 PM
The formula is

M=(4*pi^2)d^3/G*P^2, or

M=39.5*d^3/G*P^2.

Right thus far...

I wanted to find the mass of the sun, so

M= (39.5)(1.5×10¹¹ m)³
_________________________________
(6.67×10-¹¹ N×m²/kg²)(3.1536×10^7 s)

...no problem here...

M=1.33×10^35 m³
______________________________________
2.1035×10-³ N×m²×s/kg²*(3.1536×10^7 s)

You forgot the time.

Equation gives M = 2.00 x 10^30 kg.