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Musya8 [376]
3 years ago
15

Develop an algebraic relationship that describes a satellite moving in a circular orbit around a planet if the speed of the sate

llite is v, the mass of the planet is M, the mass of the satellite is m, the radius of the planet is R, and the altitude of the orbit is h. (You may include the appropriate constants as needed.) Assume that the frictional force is negligible. What will eventually happen to the satellite if the frictional (drag) force is not negligible? What could be done to compensate for a non-negligible drag force that would allow the satellite to maintain its orbit?
Physics
1 answer:
lana [24]3 years ago
5 0

Answer:

Explanation:

For this problem we must use Newton's second law where force is gravitational attraction

      F = m a

Since movement is circular, acceleration is centripetal.

      a = v2 / r

Let's replace

      G m M / r² = m v² / r

      G M r = v²

The distance r is from the center of the planet

      r = R + h

      v = √ GM / (R + h)

If the friction force is not negligible

      F - fr = m a

Where the friction force must have some functional relationship, for example

           Fr = b v + c v² +…

Suppose we are high enough for the linear term to derive the force of friction

          G m M / r - (m b v + m c v2) = m v2

          G M / r - b v = v²

We see that the solution of the problem gives lower speeds and that change over time.

To compensate for this friction force, the motors should be intermittently suspended to recover speed.

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Answer:

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Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

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From the question we have

force = 7.2 × 3 = 21.6

We have the final answer as

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Hope that helps!

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Explanation:

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A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc
ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

T=2\pi \sqrt{\frac{L}{g} }

Where T is period

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       g is acceleration due to gravity =     9.8m/s^{2}

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this L_{O}

L_{O} = 3/4 * 5.95m

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thus   T=2\pi \sqrt{\frac{L_{O} }{g} }

          T=2\pi \sqrt{\frac{4.4625 }{9.8} }

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