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IceJOKER [234]
3 years ago
6

Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth

rotates. these are called geosynchronous orbits. the radius of the earth is 6.37 * 106 m, and the altitude of a geosynchronous orbit is 3.58 * 107 m 1â22,000 miles2. what are (a) the speed and (b) the magnitude of the acceleration of a satellite in a geosynchronous orbit?

Physics
1 answer:
zalisa [80]3 years ago
7 0
Refer to the diagram shown below.

Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m

In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s

The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s

Part (a)
The tangential speed is
v = (R+h)*ω
   = (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s) 
   = 3066.7 m/s
   = 3.067 km/s

Part (b)
The centripetal acceleration is
a = v²/(R+h)
   = (3066.7 m/s)²/(4.217 x 10⁷ m)
   = 0.223 m/s²

Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²

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Answer:

a. A baseball after it has been  hit - not in free fall

b. A rock that is thrown in the  air - not in free fall

c. The moon - free-fall

d. A paper airplane - not in free fall

e. A bird flying - not in free fall

Explanation:

  1. The free-fall is defined as the falling of an object due to the action of gravity. The object is not experiencing any other force neglecting the air resistance.
  2. If an object is in free-fall, the direction of its motion is directed towards the center of the earth. It does not have a horizontal component of velocity.
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a. A baseball after it has been  hit - not in free fall according to point 1 & 2.

b. A rock that is thrown in the  air - not in free fall according to point 1.

c. The moon - free-fall according to point 3.

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e. A bird flying - not in free fall according to point 1 & 2.

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Height = 9 feet

Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²

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Average outside temperature - Wall temperature = 40 - 18 = 22 °F

Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;

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