How muich would a one hundred fifty kilogram object weigh on the surface of a lump of gold one hundred times the mass of the sun?
2007-05-09 17:59:48 · 3 answers · asked by me8md 3 in Science & Mathematics ➔ Physics
It's a block hole. It's escape velocity is greater than the speed of light so asking what the gravitation force is seems to be totally irrelevant.
2007-05-09 19:25:09 · answer #1 · answered by Gene 7 · 0⤊ 0⤋
There are many factors.
Force = G * Mass / distance ^2
-G is the gravitational constant http://en.wikipedia.org/wiki/Gravitational_constant
-Mass in this case 100 solar masses.
-Distance is how far away you are from the center of the lump of gold. In other words we need to know the radius of this lump of gold (the distance from the center to the surface)
Figure all those numbers out, and you can answer your question.
2007-05-10 01:44:58 · answer #2 · answered by Roman Soldier 5 · 0⤊ 0⤋
I will give you the method, but will not calculate for you.
1. Calculate the mass of sun (get standard data or calculate from the mass of earth - How? the centrifugal force when earth rotates = 4 *pi^2 *n^2*r where r is distance between sun and earth, n is revolution per second = 1/(365*24*3600). This force is equal to G.(mass of sun*mass of earth)/distance of sun to earth^2)
2. Once you know mass of sun, multiply it by 100 and then devide this mass by density of gold which I think is 19300 Kg/m^3. So you have the volume of gold.
3. From volume of gold find radius of lump of gold as the volume is 4*pi*r^3/3
4. You have mass of gold, the mass of object, distance (the radius of lump of gold)
5. Calculate the force by equation F= G*MASS OF GOLD*MASS OF OBJECT / (RADIUS OF GOLD)^2
gOOD LUCK
2007-05-10 02:22:48 · answer #3 · answered by dipakrashmi 4 · 0⤊ 0⤋