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Pie
3 years ago
10

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

ved by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass M.
For all parts of this problem, where appropriate, use G for the universal gravitational constant.

Required:
a. Find the kinetic energy K of a satellite with mass m in a circular orbit with radius R.
b. Find the orbital period T.
c. Find L, the magnitude of the angular momentum of the satellite with respect to the center of the planet. Express your answer in terms of m, M, G, and R.
Physics
1 answer:
ELEN [110]3 years ago
4 0

Answer:

a)  v² = G M R³,  b)  T = 2π /\sqrt{GMR}, c)  m \sqrt{GMR^5 }

Explanation:

a) The kinetic energy is

         K = ½ m v²

       

to find the velocity let's use Newton's second law

          F = m a

acceleration is centripetal

           a = v² / R

           

force is the universal force of attraction

           F = G m M / r²

we substitute

          G m M R² = m v² R

          v² = G M R³

           

the kinetic energy is

          K = ½ m G M R³

b) angular and linear velocity are related

          v = w R

          w = v / R

          w = \frac{\sqrt{GMR^3  }}{R}

          w = \sqrt{GMR}

the angular velocity is related to the period

          w = 2π / T

          T = 2π / w

we substitute

           T = 2π /\sqrt{GMR}

c) the angular moeomto is

            L = m v r

            L = m RA G M R³ R

            L =  m \sqrt{GMR^5 }

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bearhunter [10]
Sure.
Can I use your answer to part-'a' ?

If the angular acceleration is actually 32 rev/min², than
after 1.2 min, it has reached the speed of

                 (32 rev/min²) x (1.2 min)  =  38.4 rev/min .

Check:

If the initial speed is zero and the final speed is 38.4 rpm,
then the average speed during the acceleration period is

                  
(1/2) (0 + 38.4)  =  19.2 rpm  average

At an average speed of  19.2 rpm for 1.2 min,
it covers

                   (19.2 rev/min) x (1.2 min)  =  23.04 revs .

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4 0
3 years ago
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Define what is boyleâs law?
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Answer:

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Explanation:

Boyle's law: It states that the volume of given mass of gas is inversely proportional  to the pressure of gas at constant temperature.

Mathematical representation:

Suppose, a gas of mass m

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8 0
3 years ago
A ball thrown by Ginger is moving upward through the air. Diagram A shows a box with a downward arrow. Diagram B shows a box wit
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As the ball is moving in air as well as we have to neglect the friction force on it

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So the correct FBD will contain only one force and that force must be vertically downwards

So here correct answer must be

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8 0
3 years ago
In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the
Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

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8 0
3 years ago
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
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The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
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  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
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