The first known collision between space debris and a functioning satellite occurred in 1996: At an altitude of700km, a year old French spy satellite was hit by a piece of an Ariane rocket. A stabilizing boom on the satellite was demolished, and the satellite was sent spinning out of control. Just before the collision and in kilometresper hour, what was the speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was

(a) head-on and

(b) along perpendicular paths?

Short Answer

Expert verified
  1. The speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was head-on is5.4×104km/h.
  2. The speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was along perpendicular paths is 3.8×104km/h.

Step by step solution

01

Step 1: Given

The altitude is 700km.

02

Determining the concept

Using the formula for orbital speed, findthe speed of the rocket piece relative to the satellite using the given altitude and mass of the object in circular orbit. The collision was head-on and along perpendicular paths.

Formula is as follow:

v=GMR

where, v is velocity, G is gravitational constant, M is mass and R is radius.

03

(a) Determining the speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was head-on

Now,

v=GMR

But the speed of an object in a circular orbit of mass,M=5.98×1024 kgand radius,

R=(700 m+6370 km)=7070km=7.07×106 m

v=6.67×10-11Nm2kg2(5.98×1024 kg)7.07×106 m=7.51×103 m/s=2.7×104 km/h

But, for head on collision, the speed is2v, because both the objects are moving at the same speed towards each other. Hence,

v=5.4×104 km/h

Therefore, the speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was head-on is v=5.4×104 km/h.

04

(b) Determining the speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was along perpendicular paths

For perpendicular collision, the relative speed is given by Pythagorean Theorem,

v=v2+v2

v=2.7×104kmh2+2.7×104kmh2=3.8×104km/h

Hence, thespeed of the rocket piece relative to the satellite if both were in circular orbits and the collision was along perpendicular paths is3.8×104km/h.

Therefore, using the formula for orbital velocity, and the concept of relative motion and Pythagoras Theorem, the relative speed can be found.

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