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Closest distance of meteor approaching Earth

A meteor is moving at a speed of 32000 km/hr relative to the centre of Earth at a point
560 km from the surface of the Earth. At this time it has a radial velocity of 6400 km/hr.
How close does it come to the Earth's surface? (RE =6373 km)

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Solution Preview

Let's formulate this problem in more general terms: We have a meteor that is moving at some velocity v1. We know what the magnitude of the velocity, i.e. the speed |v1|, is when it is at a point 560 km above the Earth's surface:

|v1| = 32000 km/hr

Also we know the radial component of v1 at that point, which we can denote as vr1. According to the given figure in the problem, we have:

vr1 = - 6400 km/hr

at that point when the meteor is 560 km from the surface of the Earth. The minus sign is infered from the way the problem is formulated; the problem strongly suggests that the meteor is approaching Earth.

The tangential component of the velocity, vt1, at that point is then fixed as we have:

|v1|^2 = vr1^2 + vt1^2 ------------>

vt1 = 31353 km/hr

This is thus at a point 560 km above the surface of the Earth, which is at

r1 = Re + 560 km = 6933 km from the center of the Earth.

We want to find how close the meteor comes to the surface of the Earth. Obviously, at the point ...

Solution Summary

We derive the solution from first principles using conservation of energy and conservation of angular momentum.