Answer
given,
mass of satellite = 545 Kg
R = 6.4 x 10⁶ m
H = 2 x 6.4 x 10⁶ m
Mass of earth = 5.972 x 10²⁴ Kg
height above earth is equal to earth's mean radius
a) satellite's orbital velocity
centripetal force acting on satellite =
gravitational force =
equating both the above equation
v = 5578.5 m/s
b)
T = 14416.92 s
T = 4 hr
c) gravitational force acting
F = 5202 N
Answer:
Explanation:
(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...
4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity
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(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...
d = vt
d = (13.8 m/s)(2 s) = 27.6 m
The water is about 27.6 m below ground.
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* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:
vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2
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If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.