A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 x10^10 m (inside the orbit of Mercury), at which point its speed is 9.1 x10^4 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 x10^12 m from the Sun? (This is the approximate distance of Pluto from the Sun.)
To find the speed of the comet when it is 6 x 10^12 m from the Sun, we can use the concept of conservation of mechanical energy. In an elliptical orbit, the total mechanical energy remains constant.
The formula for total mechanical energy is given by:
E = (1/2)mv^2 - G(Mm)/r
Where:
E is the total mechanical energy
m is the mass of the comet
v is the velocity of the comet
G is the gravitational constant
M is the mass of the Sun
r is the distance from the Sun to the comet
Let's calculate the total mechanical energy at the closest approach of the comet to the Sun:
E1 = (1/2)m(v1)^2 - G(Mm)/r1
Given data:
v1 = 9.1 x 10^4 m/s
r1 = 4.8 x 10^10 m
Now, let's calculate the total mechanical energy at a distance of 6 x 10^12 m from the Sun:
E2 = (1/2)m(v2)^2 - G(Mm)/r2
Given data:
r2 = 6 x 10^12 m
Since the total mechanical energy remains constant:
E1 = E2
(1/2)m(v1)^2 - G(Mm)/r1 = (1/2)m(v2)^2 - G(Mm)/r2
Now, we can rearrange the equation to solve for v2 (the velocity at a distance of 6 x 10^12 m):
(v2)^2 = (v1)^2 + 2G(M/r1 - M/r2)
Substituting the given values:
(v2)^2 = (9.1 x 10^4 m/s)^2 + 2G(M/r1 - M/r2)
Now, we know G, M, r1, and r2. Plugging these values into the equation and solving for (v2)^2, we can find v2.
Using the gravitational constant G ≈ 6.67430 x 10^-11 m^3/kg/s^2, and the mass of the Sun M ≈ 1.989 x 10^30 kg, we can plug in the numbers and calculate v2.