When a spacecraft launched off the surface of the Earth needs to escape the Eart

Question

When a spacecraft launched off the surface of the Earth needs to escape the Earth’s gravity, we did not include in our calculation the effect of the Earth’s rotation to find out what its launch speed needs to be. For example, when the launch happens on the Equator in the direction of the Earth rotation, the launch speed relative to the ground is less than when it is launched against the direction of rotation. (a) Calculate the speed relative to the center of the Earth of an object at rest on the Equator. (b) Relative to the Earth’s surface, what speed must be given to a spacecraft in order for it to escape the Earth’s gravitational pull, if it launched in the direction of rotation? What if it were to be launched against the direction of rotation? (c) By what percentage is the work required to accelerate the spacecraft from rest to its launch speed relative to the Earth reduced because of the Earth’s rotation?

Explanation / Answer

If rotation of earth is not considered

let the velocity be v

0.5*m*v^2=GM*m/R

R=6400000 m

v=sqrt(2*GM/R)=11.16 km/s

if launched opposite to rotation

speed of earth rotation=460m/s

velocity be v

0.5*m*(v-460)^2=GMm/R

we get v=11.6 km/s

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