A small mailbag is released from a helicopter that is descending steadily at 2.52 m/s. (a) After 4.00 s, what is the speed of th
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Answer:
v = √ 2 G M/
Explanation:
To find the escape velocity we can use the concept of mechanical energy, where the initial point is the surface of the earth and the end point is at the maximum distance from the projectile to the Earth.
Initial
Em₀ = K + U₀
Final
=
The kinetic energy is k = ½ m v²
The gravitational potential energy is U = - G m M / r
r is the distance measured from the center of the Earth
How energy is conserved
Em₀ =
½ mv² - GmM / = -GmM / r
v² = 2 G M (1 / – 1 / r)
v = √ 2GM (1 / – 1 / r)
The escape velocity is that necessary to take the rocket to an infinite distance (r = ∞), whereby 1 /∞ = 0
v = √ 2GM /