Recall that the escape velocity is the speed needed to break free from the gravi
ID: 2871106 • Letter: R
Question
Recall that the escape velocity is the speed needed to break free from the gravitational pull of a massive body, without further propulsion.^1 The escape velocity does not take the resistance of the atmospheric air, or drag, into account. It equally ignores the rotation of the massive body. For example, the Earth's Equator stretches for 40,075 km. Our planet makes one full revolution around its axis in 24 hours. Thus a point on the Equator moves at the speed of 1,670 km/h. We can utilise the motion by launching a space vehicle in the direction of the Earth's rotation and as close to the Equator as possible.^2 Given that g = 9.8 m/s^2, find the escape velocity, in km/s, on the surface of the Earth.Explanation / Answer
Escape velocity is defined to be the minimum velocity an object must have in order to escape the gravitational field of the earth, that is, escape the earth without ever falling back.
The object must have greater energy than its gravitational binding energy to escape the earth's gravitational field. So:
1/2 mv2 = GMm/R
Where m is the mass of the object, M mass of the earth, G is the gravitational constant, R is the radius of the earth, and v is the escape velocity. It simplifies to:
v = sqrt(2GM/R)
or
v = sqrt(2gR)
Where g is acceleration of gravity on the earth's surface.
The value evaluates to be approximately: 11.1 km/s
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