Please answer part A and B. show all work please :D 7. Earth has a bad day. It will lead to a brighter future for you though, because itwill give us the Moon. Earth is hit by another planet ½ of Earth's diameter(called Theia, the Greek Goddess mother of Selene – The Goddess of the Moon).This really did happen.A) Start with how the Earth-Moon system appears today. Use the data given below,assume that both bodies are uniform spheres and calculate the angular momentum of theMoon due to its rotation, due to its orbital motion and the angular momentum of the Earthdue to its rotation. List them from largest to smallest. You may assume that the Moonorbits the center of the Earth.B) Assume that the Earth's day was originally four times longer and its original masswas 90% of its current mass. Computer models show that to get the right amount ofmaterial into orbit, the impactor had to hit in a grazing collision. Assume that the twospherical bodies barely touched at impact (this would never happen because the forcesacting on both bodies deformed them a lot – but go with it). How fast must Theia havebeen moving on impact?Useful Information1. Diameter of Earth: 12756 km.2. Diameter of the Moon: 3474 km.3. Mass of Earth: 5.98e24 kg.4. Mass of the Moon: 1/81 of the Earth.5. Current orbital period of the Moon: 27.3 days.6. Average distance from Earth to Moon: 384,000 km.

Answered: 7. Earth has a bad day. It will lead to…

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter10: Rotational Motion And Angular Momentum

Section: Chapter Questions

Problem 42PE: Consider the Earth-Moon system. Construct a problem in which you calculate the total angular...

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Please answer part A and B. show all work please :D

7. Earth has a bad day. It will lead to a brighter future for you though, because it
will give us the Moon. Earth is hit by another planet ½ of Earth's diameter
(called Theia, the Greek Goddess mother of Selene – The Goddess of the Moon).
This really did happen.
A) Start with how the Earth-Moon system appears today. Use the data given below,
assume that both bodies are uniform spheres and calculate the angular momentum of the
Moon due to its rotation, due to its orbital motion and the angular momentum of the Earth
due to its rotation. List them from largest to smallest. You may assume that the Moon
orbits the center of the Earth.
B) Assume that the Earth's day was originally four times longer and its original mass
was 90% of its current mass. Computer models show that to get the right amount of
material into orbit, the impactor had to hit in a grazing collision. Assume that the two
spherical bodies barely touched at impact (this would never happen because the forces
acting on both bodies deformed them a lot – but go with it). How fast must Theia have
been moving on impact?
Useful Information
1. Diameter of Earth: 12756 km.
2. Diameter of the Moon: 3474 km.
3. Mass of Earth: 5.98e24 kg.
4. Mass of the Moon: 1/81 of the Earth.
5. Current orbital period of the Moon: 27.3 days.
6. Average distance from Earth to Moon: 384,000 km.

Expert Solution

Step 1

Hello. Since your question has multiple parts, we will solve the first part for you. If you want the remaining part to be solved, then please resubmit the whole question and specify the part you want us to solve.

Let D_E and D_M, and R_E and R_M denote the diameters and radii of the Earth and the Moon, respectively. Let d denote the distance of the Moon from the Earth.

Let M_E and M_M denote the masses of the Earth and the Moon, respectively.

Let T_M and T_E denote the Moon’s rotation period (also its orbital period) and the Earth’s rotation period, respectively.

Let I_0M, I_M, and I_E denote the moments of inertia of the Moon about its rotational axis, that about its revolution axis, and that of the Earth about its rotational axis, respectively.

Step 2

Evaluate the Moon’s rotational angular momentum (L_0M) from its rotational angular speed (ω_0M) and its rotational period (T_M) as follows:

$\begin{array}{rcl} L_{0 M} & = & I_{0 M} ω_{0 M} \\ = & (\frac{2}{5} M_{M} R_{M}^{2}) (\frac{2 π}{T_{M}}) \\ = & \frac{4 π (\frac{M_{E}}{81}) {(\frac{D_{M}}{2})}^{2}}{5 T_{M}} \\ = & \frac{4 π (\frac{(5.98 \times 10^{24} kg)}{81}) {(\frac{(3474 km) \frac{(10^{3} m)}{(1 km)}}{2})}^{2}}{5 (27.3 days) \frac{(24 h)}{(1 days)} \frac{(3600 s)}{(1 h)}} \\ = & 6.83 \times 10^{28} kg \cdot m^{2} / s \end{array}$

Step 3

Assume that the Moon’s center is at the distance d from the Earth’s center.

First, determine I_M by using the parallel axis theorem, and then determine the Moon’s orbital angular momentum (L_M) from its orbital angular speed (ω_M) and its orbital period (T_M) as follows:

$\begin{array}{rcl} L_{M} & = & I_{M} ω_{M} \\ = & (I_{0} + M_{M} d^{2}) (\frac{2 π}{T_{M}}) \\ = & \frac{2 π (\frac{M_{E}}{81} (\frac{2}{5} {(\frac{D_{M}}{2})}^{2} + d^{2}))}{T_{M}} \\ = & \frac{2 π (\frac{(5.98 \times 10^{24} kg)}{81}) (\frac{{((3474 km) \frac{(10^{3} m)}{(1 km)})}^{2}}{10} + {((384000 km) \frac{(10^{3} m)}{(1 km)})}^{2})}{(27.3 days) \frac{(24 h)}{(1 days)} \frac{(3600 s)}{(1 h)}} \\ = & 2.35 \times 10^{36} kg \cdot m^{2} / s \end{array}$

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