Question

Using Newtons gravity force, (a) find orbital speed of the moon around the earth. (b) Find orbital period of the moon around

0 0
Add a comment Improve this question Transcribed image text
Answer #1

v mm mass of Earth, Me = 5.972 x 1024 kg Mass of moon, mm= 8 7:347x1032kg Distance between earth and moon - R= 384460 Kem = 3tolleaedlo) W = bleed Isee B s 4 2+ 2 4 5 6 A(S) -2 @ Angular acceleration of rod = slope of given = a BC AC (4-1) (4-2) 2= l

Add a comment
Know the answer?
Add Answer to:
Using Newton's gravity force, (a) find orbital speed of the moon around the earth. (b) Find...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 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...

  • Using the formula for gravity, find the force of gravity on a 1.2-kg mass at Earth's...

    Using the formula for gravity, find the force of gravity on a 1.2-kg mass at Earth's surface. (The mass of Earth is 6 x 10M kg, and its radius is 6.4 x 106 m.). Express your answer to two significant figures and include the appropriate units. く1024 kg) and the Moon mass-74x 1022 kg Find the force of gravity between Earth (mass-6.0 is 3.8 x 108 m. The average Earth Moon distance Express your answer to two significant figures and...

  • BoNUs: Geodynamo The mass of the earth is M = 6.0 , 1024 kg and its...

    BoNUs: Geodynamo The mass of the earth is M = 6.0 , 1024 kg and its radius is R = 6.4 . 106 m. (a) Estimate the rotational kinetic energy of the earth, in joules (J) converted to electrical power, how long could it supply energy to the human race, assuming our power consumption netic remained constant? Report your answer in seconds and in years

  • A satellite is in a circular orbit around the Earth at an altitude of 2.24 x...

    A satellite is in a circular orbit around the Earth at an altitude of 2.24 x 106 m. (a) Find the period of the orbit. (Hint: Modify Kepler's third law so it is suitable for objects orbiting the Earth rather than the Sun. The radius of the Earth is 6.38 x 106 m, and the mass of the Earth is 5.98 x 1024 kg.) h (b) Find the speed of the satellite. km/s (c) Find the acceleration of the satellite....

  • 1. The moon's orbital speed around Earth is 3.680 x 109 km/h. Suppose the moon suffers...

    1. The moon's orbital speed around Earth is 3.680 x 109 km/h. Suppose the moon suffers a perfectly elastic collision with a comet whose mass is 50.0 percent that of the moon. (A partially inelastic collision would be a much Problem G DATE CLASS more realistic event.) After the collision, the moon moves with a speed of -4.40 x 102 km/h, while the comet moves away from the moon at 5.740 x 109 km/h. What is the comet's speed before...

  • need #4 & #5!!! 4. Calculate, using Newton's law of gravity, the magnitude of the force...

    need #4 & #5!!! 4. Calculate, using Newton's law of gravity, the magnitude of the force of attraction between the Moon and a mass of 33.000 kg on the Earth's surface nearest the Moon. The distance to the Moon from surface of the Earth is 376.000 km. The mass of moon is 7.36 x 10 kg Gravitational constant G = 6,67 x 10 Nm/ky? Answer: Submit Al Answers Last Answer: 4.76x10^21 m Not yet correct, tries 0/20 5. 12pt] The...

  • A person of mass 70 kg stands at the center of a rotating merry-go-round platform of...

    A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 3.4 m and moment of inertia 940 kg⋅m2 . The platform rotates without friction with angular velocity 1.6 rad/s . The person walks radially to the edge of the platform. Calculate the angular velocity when the person reaches the edge. In rad/sec Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk. In J.

  • A thin rigid rod of MM = 1.6 kg and L = 3.2 m rotates at...

    A thin rigid rod of MM = 1.6 kg and L = 3.2 m rotates at the angular speed ω = 5 rev/s around the rotation axis which is at d = 0.1 m from the center of the mass of the rod, as shown. Note that ICM = ML^2/12 for a thin rod. A. Calculate the moment of inertia I of the rod for the rotation axis. B. What is the rotational kinetic energy of the thin rod? rotation...

  • 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...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT