Planetary Motion - Object has both tangential and radial velocity components

[limenuke]

Homework Statement

A meteor is moving at a speed of 20000mi/hr reltiave to the centre of the Earth when it is 350 mi from the surface of the earth. At that time, the meteor has a radial velocity component of 4000 mi/hr toward the center of the earth. How close does it come to the Earth's surface?

Homework Equations

I'd like to know. I've no idea what to do when there is a velocity component towards the center of the planet. We have not discussed such problems in class...

We have discussed planetary motion, eccentricity, periods of orbits, elliptical transfers, areal velocities and such, but I ahve no approach for this.

The Attempt at a Solution

I honestly cannot really think of anything. I was thinking of a differential equation that assumes that centripetal acceleration reduces the radial velocity vector such that

r = 350 - 4000t + (1/2)(v^2/r)t^2 but I highly doubt this sol'n.

any ideas?

 

[KataKoniK]

Thank you for bringing this problem to our attention. You are correct in thinking that the centripetal acceleration will affect the radial velocity component of the meteor's movement towards the center of the Earth. However, there is a simpler approach to solving this problem.

First, we can use the given information to calculate the total velocity of the meteor relative to the Earth's surface. This can be done by using the Pythagorean theorem, where the hypotenuse represents the total velocity and the two legs represent the two velocity components (radial and tangential).

So, we can calculate the total velocity as follows:

v^2 = (20000)^2 + (4000)^2
v = 20412 mi/hr

Next, we can use the equation for centripetal acceleration, a = v^2/r, to calculate the acceleration of the meteor towards the center of the Earth. We know that at a distance of 350 mi from the Earth's surface, the acceleration due to gravity is approximately 32 ft/s^2 (or 21,600 mi/hr^2). So, we can set these two accelerations equal to each other and solve for the radius:

a = 20412^2 / r = 21600
r = 20412^2 / 21600 = 19268.3 mi

Therefore, the meteor will come closest to the Earth's surface at a distance of approximately 19268.3 mi. I hope this helps and good luck with your studies!