# Closest distance of meteor approaching Earth

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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 Summary

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

##### 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 ...

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