An asteroid in an elliptical orbit about the sun travels at 2.0 106 m/s at perihelion (the point of closest approach) at a distance of 3.8 108 km from the sun. How fast is it traveling at aphelion (the most distant point), which is 8.3 108 km from the sun?
the angular momentum of the system is constant, so the velocity is inversely proportional the the radius of the orbit
Va = Vp * (Rp / Ra)
To find the speed of the asteroid at aphelion, we can use the concept of conservation of angular momentum. According to this principle, the product of an object's mass, velocity, and the radius of its orbit remains constant as long as there is no external torque acting on it.
Given that the speed of the asteroid at perihelion is 2.0 106 m/s and its distance from the sun is 3.8 108 km (which can be converted to 3.8 × 1011 m), we can use these values to find the angular momentum.
Angular momentum (L) = mass (m) × velocity (v) × radius (r)
L = m × v × r
Now, to find the mass of the asteroid, we need to know its mass in order to calculate the speed at aphelion. However, the mass of the asteroid is not given in the question. Therefore, we need to make an assumption. Let's assume the mass of the asteroid is m kg.
Now, we have:
L1 = m × v1 × r1
where L1 is the angular momentum at perihelion.
To solve for m, we can rearrange the equation:
m = L1 / (v1 × r1)
Next, we can use this value of m to find the speed at aphelion (v2).
L2 = m × v2 × r2
where L2 is the angular momentum at aphelion, r2 is the distance at aphelion (8.3 × 108 km = 8.3 × 1011 m), and v2 is the speed at aphelion (which we want to find).
By substituting the value of m, we get:
L2 = (L1 / (v1 × r1)) × v2 × r2
Now, we can rearrange and solve for v2:
v2 = (L2 × v1 × r1) / (L1 × r2)
Substituting the given values:
L1 = (m × v1 × r1) = (m × 2.0 × 106 m/s × 3.8 × 1011 m)
L2 = (m × v2 × r2) = (m × v2 × 8.3 × 1011 m)
Now we can substitute these values in the equation:
v2 = (m × v2 × 8.3 × 1011 m) / (m × 2.0 × 106 m/s × 3.8 × 1011 m)
Simplifying, we can cancel out the masses and distance units:
v2 = (v2 × 8.3 × 1011 m) / (2.0 × 106 m/s × 3.8 × 1011 m)
v2 = (8.3 / (2.0 × 3.8)) × 1011 / 106 m/s
Let's solve this equation:
v2 ≈ 2.18 × 103 m/s
Therefore, the asteroid is traveling at approximately 2.18 × 103 m/s at aphelion.