Uranus is approximately 2.74 x 10^9 km away from the Sun at perihelion and 3.00 x 10^9 km at aphelion. Find the eccentricity of Uranus's orbit given that the perihelion is r = a(1 - e) and the aphelion is r = a(1 + e).
so, plug and chug.
a/(1-e) = 3.00*10^9
a/(1+e) = 2.74*10^9
a = 2.86*10^9
e = 0.0453
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To find the eccentricity of Uranus's orbit, we can use the given information about the distances at perihelion and aphelion. Let's assume that "a" represents the average distance between Uranus and the Sun.
We can start by using the equation for perihelion:
r = a(1 - e)
Substituting the given perihelion distance of 2.74 x 10^9 km:
2.74 x 10^9 = a(1 - e)
Similarly, we can use the equation for aphelion:
r = a(1 + e)
Substituting the given aphelion distance of 3.00 x 10^9 km:
3.00 x 10^9 = a(1 + e)
Now we have a system of two equations:
2.74 x 10^9 = a(1 - e)
3.00 x 10^9 = a(1 + e)
Now, let's solve this system of equations to find the value of "e".
We can start by multiplying both sides of the second equation by (1 - e) and the first equation by (1 + e):
2.74 x 10^9(1 + e) = a(1 - e)(1 + e)
3.00 x 10^9(1 - e) = a(1 + e)(1 - e)
Expanding and simplifying both equations:
2.74 x 10^9 + 2.74 x 10^9e = a(1 - e^2)
3.00 x 10^9 - 3.00 x 10^9e = a(1 - e^2)
Now, we can subtract the second equation from the first:
2.74 x 10^9 + 2.74 x 10^9e - (3.00 x 10^9 - 3.00 x 10^9e) = a(1 - e^2) - a(1 - e^2)
Simplifying the equation:
2.74 x 10^9 - 3.00 x 10^9 + (2.74 x 10^9e + 3.00 x 10^9e) = 0
Combining like terms:
-0.26 x 10^9 + 5.74 x 10^9e = 0
Dividing both sides by 5.74 x 10^9:
e = -(-0.26 x 10^9) / 5.74 x 10^9
e ≈ 0.045
Therefore, the eccentricity of Uranus's orbit is approximately 0.045.
To find the eccentricity (e) of Uranus's orbit, we can use the given information about the distances at perihelion (r = 2.74 x 10^9 km) and aphelion (r = 3.00 x 10^9 km).
The perihelion distance, r = a(1 - e), where "a" represents the semi-major axis of the orbit. We can rearrange this equation to solve for e:
r = a(1 - e)
r/a = 1 - e
e = 1 - (r/a)
Similarly, for the aphelion distance, r = a(1 + e), we can rearrange the equation to solve for e:
r = a(1 + e)
r/a = 1 + e
e = (r/a) - 1
Now, let's substitute the given distances into the equations to find the eccentricity:
At perihelion:
e = 1 - (r/a)
e = 1 - (2.74 x 10^9 km / a)
At aphelion:
e = (r/a) - 1
e = (3.00 x 10^9 km / a) - 1
Since we have the distances at both perihelion and aphelion, we can set them equal to each other and solve for "a":
2.74 x 10^9 km = a(1 - e)
3.00 x 10^9 km = a(1 + e)
Solving these two equations simultaneously will allow us to find the value of "a".
Once we have the value of "a", we can substitute it back into either of the equations for e, and solve for the eccentricity.