Question

Neptune has a mass of 1.0×1026 kg and is 4.5×109 km from the Sun with an orbital period of 155 years. Planetesimals in the outer primordial solar system 4.5 billion years ago coalesced into Neptune over hundreds of millions of years. If the primordial disk that evolved into our present day solar system had a radius of 1011 km and if the matter that made up these planetesimals that later became Neptune was spread out evenly on the edges of it, what was the orbital period of the outer edges of the primordial disk? Express your answer rounded to the nearest year, and be sure not to do any rounding before expressing your final answer.

P=___years
A playground merry-go-round has a mass of 140 kg and a radius of 1.8 m and it is rotating with an angular velocity of 0.6rev/s. What is its angular velocity after a 22−kg child gets onto it by grabbing its outer edge? The child is initially at rest.___ rev/s

Answer 1

The orbital period of the outer edges of the primordial disk was approximately 620 years.

Step-by-step explanation:

To determine the orbital period of the outer edges of the primordial disk, we can use the concept of angular momentum. The angular momentum of an object is proportional to the square of its size (diameter) divided by its period of rotation. According to the given information, the primordial disk had a radius of 1011 km. If the matter that later became Neptune was spread out evenly on the edges of the disk, the final diameter would be twice the radius or 2x1011 km. This means the final diameter is 2 times the initial diameter.

Since angular momentum is conserved, we can use the concept of angular momentum conservation to find the orbital period of the outer edges of the primordial disk. The initial period of rotation is given as 155 years for Neptune. We can set up the equation D1²/P1 = D2²/P2, where D1 and D2 are the initial and final diameters respectively, and P1 and P2 are the initial and final periods of rotation. Substituting the known values, we have (1011 km)² / 155 years = (2(1011 km))² / P2. Solving for P2 gives us a value of approximately 620 years. Therefore, the orbital period of the outer edges of the primordial disk was approximately 620 years.

Answer 2

The orbital period of the outer edges of the primordial disk can be estimated using Kepler's third law and the conservation of angular momentum, although a precise calculation would require a more complex astronomical model. The new angular velocity of a playground merry-go-round after a child gets on it can be calculated using conservation of angular momentum.

Step-by-step explanation:

To determine the orbital period of the outer edges of the primordial disk, we apply Kepler's third law, which states that the square of the orbital period (P) is proportional to the cube of the semi-major axis of its orbit (a), which can be represented as P² ≈ a³. To simplify the calculation process, we consider another example from the primordial solar system where the initial diameter was 10,000 AU and the rotation period was 1,000,000 years; when shrunk to the size of Pluto's orbit, which has a diameter of about 80 AU, the new rotation period became 64 years. From this, we can establish that if the size decreases by a certain factor, the period decreases by the cube root of that same factor.

However, the actual calculation to find the orbital period of the material at the edge of the primordial disk with a radius of 10¹¹ km is a complex problem requiring more specific identification of the exact dynamics involved in the solar nebula, which typically involves detailed astronomical models and computations.

For the playground merry-go-round question, conservation of angular momentum must be used, since no external torques are acting on the system. The initial angular momentum can be calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. The moment of inertia for a solid disk is I = (1/2)MR². When the child gets on the merry-go-round, the new moment of inertia is the sum of the initial moment of inertia and the moment of inertia of the child considered as a point mass at the edge (I_child = MR²). Therefore, the final angular velocity ω_final can be found by equating the initial angular momentum to the final angular momentum (L_initial = L_final), solving for ω_final. The angular velocity is typically expressed in revolutions per second (rev/s).