The distance between Jupiter and the sun is 5.2 AU.
According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.
Now, we are told that the orbital period (T) is 11. 9 Earth years. We have to make the distance the subject of the formula.
r =T^2/3
r = (11.9)^2/3
r = 5.2 AU
Learn more: brainly.com/question/15207516
I believe the answer is Ca
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:
-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.
-- Then the effort needed to lift the load is
(the weight of the load) x (13 / the distance between the fulcrum and the effort)
Jake will start to be addicted to his computer and getting no exercise and he will become lazy
