where it is 7.3 x 10^9 km from the sun
Pluto moves in a fairly elliptical orbit around the sun. Pluto's speed at its closest approach of 4.43 x 10^9 km is 6.12 km/s
Pluto moves in a fairly elliptical orbit around the sun. Pluto's speed at its closest approach of 4.43 x 10^9 km is 6.12 km/s
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The easiest way to answer this question is that angular momentum is conserved. So L at the closest is equal to L at the furthest point.
L is equal to mvr, where m is the mass, v is the velocity, and r is the radius.
So if angular momentum is conserved, then mv1r1 = mv2r2. The mass cancels, leaving v1r1 = v2r2.
So if it is at it's furthest point, then velocity needs to be smaller in order to compensate and keep angular momentum constant, and then when r gets to it's lowest point, then v would be at it's highest, with the same argument from above.
L is equal to mvr, where m is the mass, v is the velocity, and r is the radius.
So if angular momentum is conserved, then mv1r1 = mv2r2. The mass cancels, leaving v1r1 = v2r2.
So if it is at it's furthest point, then velocity needs to be smaller in order to compensate and keep angular momentum constant, and then when r gets to it's lowest point, then v would be at it's highest, with the same argument from above.
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The force which causes Pluto to orbit at a specific velocity, at a specific distance, is the Universal force of gravitational attraction. The equation is shown below.
Fg = (G * M1 * M2) ÷ d^2
G = 6.67 * 10^-3
M1 = mass of sun
M2 = mass of Pluto
d = distance between the sun and Pluto
(G * M1 * M2) is constant, so Fg = K ÷ d^2
This force is inversely proportional the square of the distance.
If the distance is doubled, the force will divided by 4.
When considering the velocity, we say that this force is equal to the centripetal force.
Fg = Fc
Fc = m * v^2/r
m = mass of Pluto
v = velocity
r = distance between the sun and Pluto = d
Fc = M2 * v^2 ÷ d
Set Fc = Fg
M2 * v^2 ÷ d = (G * M1 * M2) ÷ d^2
Divide both sides by M2
v^2 ÷ d = (G * M1) ÷ d^2
Multiply both sides by d
v^2 = (G * M1) ÷ d
Velocity = [(G * M1) ÷ d]^0.5
Velocity = (G * M1)^0.5 ÷ d^0.5
Velocity * d^0.5 = (G * M1)^0.5 = constant
According to the equation above, the velocity is inversely proportional to the square root of the distance between the sun and Pluto.
v1 * √d1 = v2 * √d2
6.12 * (4.43 * 10^9)^0.5 = v2 * (7.3 * 10^9)^0.5
v2 = 6.12 * (4.43 * 10^9)^0.5 ÷ (7.3 * 10^9 )^0.5 = 6.12 * 0.779
v2 ≈ 4.77 km/s
The velocity at a greater distance is less that the velocity at a closer distance.
Fg = (G * M1 * M2) ÷ d^2
G = 6.67 * 10^-3
M1 = mass of sun
M2 = mass of Pluto
d = distance between the sun and Pluto
(G * M1 * M2) is constant, so Fg = K ÷ d^2
This force is inversely proportional the square of the distance.
If the distance is doubled, the force will divided by 4.
When considering the velocity, we say that this force is equal to the centripetal force.
Fg = Fc
Fc = m * v^2/r
m = mass of Pluto
v = velocity
r = distance between the sun and Pluto = d
Fc = M2 * v^2 ÷ d
Set Fc = Fg
M2 * v^2 ÷ d = (G * M1 * M2) ÷ d^2
Divide both sides by M2
v^2 ÷ d = (G * M1) ÷ d^2
Multiply both sides by d
v^2 = (G * M1) ÷ d
Velocity = [(G * M1) ÷ d]^0.5
Velocity = (G * M1)^0.5 ÷ d^0.5
Velocity * d^0.5 = (G * M1)^0.5 = constant
According to the equation above, the velocity is inversely proportional to the square root of the distance between the sun and Pluto.
v1 * √d1 = v2 * √d2
6.12 * (4.43 * 10^9)^0.5 = v2 * (7.3 * 10^9)^0.5
v2 = 6.12 * (4.43 * 10^9)^0.5 ÷ (7.3 * 10^9 )^0.5 = 6.12 * 0.779
v2 ≈ 4.77 km/s
The velocity at a greater distance is less that the velocity at a closer distance.
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Yeah, I'll just blank that out till I get it right.
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