Step-by-step explanation:
400-kg satellite has been placed in a circular orbit 1500 km above the surface of the earth. The acceleration of gravity at this elevation is 6.43 m/s^2s2. Determine the linear momentum of the satellite, knowing that its orbital speed is 25.6\times 10^325.6×103 km/h.
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Solution:
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The acceleration due to gravity and the height do not matter when calculating the linear momentum.
Convert the orbital speed from km/h to m/s.
v=25.6\times 10^3v=25.6×103 km/h =7111.11=7111.11 m/s
We can use the following formula to figure out the linear momentum.
L=vmL=vm
(Where LL is linear momentum, vv is velocity, and mm is mass)
L=(7111.11)(400)L=(7111.11)(400)
L=2.84\times 10^6L=2.84×106 kg·m/s
Final Answer:
L=2.84\times 10^6L=2.84×106 kg·m/s
This question can be found in Vector Mechanics for Engineers: Statics and Dynamics, 11th edition, chapter 12, question 12.3.
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