The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
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Speed of the satellite</h3>
v = √GM/r
where;
- M is mass of Earth
- G is universal gravitation constant
- r is distance from center of Earth = Radius of earth + 4930 km
v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]
v = 5,916.36 m/s
Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
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60 Miles per hour 60 times 2 is 120
Answer:
∆p=(m2v)kg.m/s
Explanation:
∆p=mv where v=2v. hence ∆p=m2v
Answer:
The gain in velocity is 0.37m/s
Explanation:
We need solve this problem though the conservation of momentum. That is,


Using the equation to find
,

Using the conservation of energy equation, we have,




Now this energy over the cannonball



The gain in velocity is 0.37m/s