between the earth and the sun to that between Jupiter and the sun. The mass of Jupiter is about 318 times the mass of the Earth. The gravitaional attraction between the sun and Jupiter is stronger by a factor of about____________?
PLEASE HELP!!
2007-10-09 08:57:43 · 5 answers · asked by f16pilottmo 2 in Science & Mathematics ➔ Astronomy & Space
The gravitational force between two objects is defined by the equation:
F = G*(m1*m2)/r²
where m1 = mass of object 1
m2 = mass of object 2
r = radius between the centers of mass of the two objects
G = gravitational constant = 6.672x10^-11 m³/kg-sec²
The value of r for the Earth-Sun system = 1.496 x 10^11 m = Res
The value of r for the Jupiter-Sun system = 7.783 x 10^11 m = Rjs
Mass of the Earth = 5.9742 x 10^24 kg = Me
Mass of Jupiter = 1.899 x 10^27 kg = Mj
Mass of the Sun = 1.9891 x 10^30 kg = Ms
So the ratio of Jupiter-Sun/Earth-Sun =
Ratio = [G*(Mj*Ms)/Rjs²]/[G* (Me*Ms)/Res²]
Simplifying, we find the G cancels out, as well as Ms, and then the equation simplifies to:
Ratio = (Mj*Res²)/(Me*Rjs²)
Substituting in our values from above, we get:
Ratio = [1.899 x 10^27 kg * (1.496 x 10^11 m)²]/
[5.9742 x 10^24 kg * (7.783 x 10^11 m)²]
Ratio = 11.74
That uses actual numbers from the reference. If we use the mass ratio given in the problem, let's see what turns up:
Ratio = 318 * (1.496 x 10^11 m)²/(7.783 x 10^11 m)²
= 11.75
As noted above, Jupiter is ~5 times farther from the Sun -- a more precise figure from the reference is 5.20283, so let's plug that in:
Ratio = 318/(5.20283)² = 11.75
Finally, the actual ratio of the masses is 317.83, so plugging that in to the equation immediately above gives:
Ratio = 317.83/(5.20283)² = 11.74
So, the gravitational force between Jupiter and the Sun is nearly twelve times that of the force between the Earth and Sun.
2007-10-09 09:35:08 · answer #1 · answered by Dave_Stark 7 · 0⤊ 0⤋
Jupiter is ~5 times more distant from the sun than Earth is so you need to use the inverse square law and divide the mass ratio by 25 to get factor you require.
2007-10-09 09:10:03 · answer #2 · answered by jim m 5 · 1⤊ 0⤋
mass of the earth ( x ) time 318 = Factor of difference ?
2007-10-09 09:02:03 · answer #3 · answered by Anonymous · 0⤊ 2⤋
(i) BY F = mg = m x GMe/Re^2-----------------(i) =>F = [6.67 x 10^-11 x 6 x 10^24 x 27.5]/[6.4 x 10^6]^2 =>F = 268.69 N (ii)On Moon:- =>F = {m x G x (Me/81.3)}/{0.27Re)^2 =>F = 1/5.93 x m x GMe/Re^2 =>F = 1/5.93 x 268.69 =>F = 45.33 N
2016-05-20 00:14:36 · answer #4 · answered by kendra 3 · 0⤊ 0⤋
63.6
Edit: actually i think its alot more than that.
im working on it.
2007-10-09 09:15:39 · answer #5 · answered by AlCapone 5 · 0⤊ 1⤋