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3 years ago

6

A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.97 x 10^4 s. What is

the speed at which the satellite travels?

1 answer:

Zepler [3.9K]3 years ago

5 0

Answer:

v = 5030.4 m/s

Explanation:

let Fc be the centripetal force and Fg be the gravitational force of attraction on the satelite. let m be the mass of the satelite and M be the mass of earth.

At any point in the orbit, the satelite experiences forces , Fc and Fg such that:

Fc = Fg

m×v^2/r = G×M×m/(r^2)

m and r are the same throughout the equation,so that:

v^2 = G×M/(r)

v = \sqrt{G×M/(r)}

In a circular orbit, the displacement the satelite should cover is 2×π×r in a period :

T = 2×π×r /v

so that:

v = 2×π×r /T

then

2×π×r /T = \sqrt{G×M/(r)}

r = \sqrt[3]{T^2×G×M/(4×π^2)}

r = \sqrt[3]{(1.97×10^4)^2×(6.67408×10^-11)×(5.98×10^24)/(4×π^2)}

r = 15772060.56 m

then :

v = 2×π×r /T

= 2×π×(15772060.56) /(1.97×10^4)

= 5030.4 m/s

therefore, the satelite travels at a speed of 5030.4 m/s.

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Since the mechanical energy must be conserved, we can write
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<span>from which we find the maximum speed
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<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span> $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
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3 years ago