What is the angular momentum of the moon about the Earth? The mass of the moon...
What is the angular momentum of the moon about the Earth? The mass of the
moon is 7.35 x 10^22 kg, the center-to-center separation of the Earth and the moon
is 3.84 x 10^5 km, and the orbital period of the moon is 27.3 days.
How do you get 3.84 x 10^8?
v = 2 x pi x r/T
L = m x 2 x pi x r x r/ T
L = (7.35 x 10^22 x 2 x 3.14 x 3.84 x 10^8 x 3.84 x 10^8) /
(27.3 x 24 x 3600)
L = 2.885 x 10^34 kgm^2/s
Homework Answers
Add Answer to:
What is the angular momentum of the moon about the Earth? The
mass of the
moon...
There is a moon orbiting an Earth-like planet. The mass of the moon is 5.37 ×...
There is a moon orbiting an Earth-like planet. The mass of the moon is 5.37 × 1022 kg, the center-to-center separation of the planet and the moon is 5.34 × 105 km, the orbital period of the moon is 21.6 days, and the radius of the moon is 1060 km. What is the angular momentum of the moon about the planet? Answer in units of kg m2 /s.
Kr. m M The center of mass of the Earth-Moon two-body system lies somewhere along the...
Kr. m M The center of mass of the Earth-Moon two-body system lies somewhere along the imaginary line that connects their centers. Find the location of center of mass of the Earth-Moon system using the data in the table, and given that r, the distance between their centers, on average, is 3.84 x 10 km. 24 Earth mass = 5.98 x 10 kg radius = 6370 km Moon mass = 7.35 x 10 kg radius = 1737 km Select one:...
The mass of Earth is 5.97 × 1024 kg, the mass of the Moon is 7.35...
The mass of Earth is 5.97 × 1024 kg, the mass of the Moon is 7.35 × 1022 kg, and the mean distance of the Moon from the center of Earth is 3.84 × 105 km. Use these data to calculate the magnitude of the gravitational force exerted by Earth on the Moon.
What centripetal force is exerted on the Moon as it orbits about Earth at a center-to-center...
What centripetal force is exerted on the Moon as it orbits about Earth at a center-to-center distance of 3.84 × 10 8 m with a period of 27.4 days? What is the source of the force? The mass of the Moon is equal to 7.35 × 10 22 kg. Please draw a free body diagram too.
The Center of Mass of the Earth-Moon-Sun System
In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate the center of mass of the earth-Moon-Sun system. The mass of the Moon is \(7.35 \times 10^{22} \mathrm{~kg}\), the mass of the Earth is \(6.00 \times 10^{24} \mathrm{~kg}\), and the mass of the sun is \(2.00 \times 10^{30} \mathrm{~kg}\). The distance between the Moon and the Earth is \(3.80 \times 10^{5} \mathrm{~km}\). The distance between the Earth and...
A moon of mass m orbits around a non-rotating planet of mass M with orbital angular velocity . The moon also rotates about its own axis with angular velocity .
1. A moon of mass \(m\) orbits around a non-rotating planet of mass \(M\) with orbital angular velocity \(\Omega\). The moon also rotates about its own axis with angular velocity \(\omega\). The axis of rotation of the moon is perpendicular to the plane of the orbit. Let \(I\) be the moment of inertia of the moon about its own axis. You can assume \(m<<M\)so that the center ofmass of the system is at the center of the planet.(a) What is...
even though the solutions are literally there i am confused and do not know how to do any of these problems ngular Momentum 2. What is the angular momentum of the moon as it revolves around the E...
even though the solutions are literally there i am confused
and do not know how to do any of these problems
ngular Momentum 2. What is the angular momentum of the moon as it revolves around the Earth? Assume the period of the moon's orbit is 27.3 days. Answer: 2.89 × 1034 kg m2/s The position of a particle of mass m with respect to the origin is given by 22. Use r x p to find an expression for...
The tidal forces between the Earth and the Moon slowed down the Moon’s rotation about its own axis until the rotation period became equal to the Moon’s orbital period around the Earth as we observe today.
The tidal forces between the Earth and the Moon slowed down the Moon's rotation about its own axis until the rotation period became equal to the Moon's orbital period around the Earth as we observe today. The same effect is also slowing down the Earth's rotation about its own axis and increasing the separation \(D\) between the Moon and the Earth at a rate of \(\Delta D / \Delta t=3.8 \mathrm{~cm}\) per year. In this problem, you can ignore the...
(a) Calculate the angular momentum (in kgm/s) of Neptune in its orbit around the Sun (The...
(a) Calculate the angular momentum (in kgm/s) of Neptune in its orbit around the Sun (The mass of Neptune is 1.020 x 10% ko, the orbital radius is 4.497 x 10 km and the orbital period is 165 y.) kg. m/s (6) Compare this angular momentum with the angular momentum of Neptune on its axis. (The radius of Neptune is 2.476 x 10 km and the rotation period is 16.11 h.) Lorbital - rotation
The Moon orbits the Earth in an approximately circular path. The position of the Moon as...
The Moon orbits the Earth in an approximately circular path. The position of the Moon as a function of time is given by x(t) = r cos(ωt) y(t) = r sin(ωt), where r = 3.84 x 10^8 m and ω = 2.46 x 10^-6 radians/s. What is the average velocity of the Moon measured over the interval from t = 0 to t = 4.68 days? Find its magnitude, in m/s, and find its direction, given as an angle measured...
Kr. m M The center of mass of the Earth-Moon two-body system lies somewhere along the imaginary line that connects their centers. Find the location of center of mass of the Earth-Moon system using the data in the table, and given that r, the distance between their centers, on average, is 3.84 x 10 km. 24 Earth mass = 5.98 x 10 kg radius = 6370 km Moon mass = 7.35 x 10 kg radius = 1737 km Select one:...
even though the solutions are literally there i am confused
and do not know how to do any of these problems
ngular Momentum 2. What is the angular momentum of the moon as it revolves around the Earth? Assume the period of the moon's orbit is 27.3 days. Answer: 2.89 × 1034 kg m2/s The position of a particle of mass m with respect to the origin is given by 22. Use r x p to find an expression for...
(a) Calculate the angular momentum (in kgm/s) of Neptune in its orbit around the Sun (The mass of Neptune is 1.020 x 10% ko, the orbital radius is 4.497 x 10 km and the orbital period is 165 y.) kg. m/s (6) Compare this angular momentum with the angular momentum of Neptune on its axis. (The radius of Neptune is 2.476 x 10 km and the rotation period is 16.11 h.) Lorbital - rotation
Most questions answered within 3 hours.
-
Where is the error in this code sequence?
String s1 = "Hello";
String s2 = "ello";...
asked 10 months ago -
Financial data for Joel de Paris, Inc., for last year
follow:
Joel de Paris, Inc.
Balance...
asked 10 months ago -
Consider this reaction:
Al2(SO4)3 (aq)+ BaCl3
(aq) Al2Cl6 (aq)- +
3BaSO4(s) . What is the...
asked 10 months ago -
Suppose that Savneet is considering increasing her
recent random sample from 20 car rentals to 40...
asked 10 months ago -
Trucks arrive at an unloading terminal at an average rate of 120
per hour.
Trucks arrive...
asked 10 months ago -
Why are methanol and ethanol completely soluble in water while
octanol is not very little soluble....
asked 10 months ago -
A facilities manager at a university reads in a research report
that the mean amount of...
asked 10 months ago -
When the CuSO4 is rehydrated by adding water to the anhydrous
compound, is this an endothermic...
asked 10 months ago -
A ray of sunlight is passing from diamond into crown glass; the
angle of incidence is...
asked 10 months ago -
A block of mass 0.249 kg is placed on top of a light, vertical
spring of...
asked 10 months ago -
how do the kidneys compensate in the presences of acidosis
a) trigger hyperventilate
b) reserve acid...
asked 10 months ago -
Question 501 pts
The rental rate of capital to the firm increases. Which of the
following...
asked 10 months ago