Answer to Question #136276 in Astronomy | Astrophysics for Harsh Deep Rautela

Question

Imagine that you are measuring the location of two objects using cylindrical polar coordinates. The first is at r=10.32 , θ=−0.32 and z=45.3 . The second is at r=10.33 , θ=−0.31 and z=45.27 .

How far apart are these two points?

Answer 1

Solution

We are given the cylindrical coordinates. we have to convert these to regular coordinates of the cartesian plane as shown below;

First point;

to find $x$ , we apply $x= r × cos(\theta)$

$= 10.32× cos(-0.32)$

$=9.7961$

to find $y$ , we apply $y = r× sin(\theta)$ `

$= 10.32× sin(-0.32)$

$= -3.2463$

And $z$ remains the same i.e. $z= 45.3$

Now the first point $p_1 = (9.7961, -32463, 45.3)$

Second point;

Finding $x$ ; $x= r × cos(\theta)$

$= 10.33× cos (-0.31)$

$= 9.8376$

`

Finding $y$ ; $y = r× sin(\theta)$

$= 10.33 × sin (-0.31)$

$=-3.1513$

And $z$ remains the same i.e. $z= 45.27$

The second point, $p_2 = (9.8376, -3.1513, 45.27)$

Now having the position of the two objects, we find the distance between them; which is given by;

$D^2 = X^2 + Y^2 + Z^2$

To find the $X, Y, and, Z$ , we Calculate the positive difference between each coordinates, i.e

$X= 9.8376-9.7961$

$=0.0415$

$Y = -3.1513 - -3.2463$

$=0.095$

$Z = 45.3-45.27$

$=0.03$

Now having the values, we calculate;

$D^2 = X^2 + Y^2 + Z^2$

$= 0.0415^2 + 0.095^2 +0.03^2$

$= 0.00172225 + 0.009025 +0.0009$

$= 0.01164725$

$D = \sqrt{ 0.01164725}$

$= 0.1079$

Therefore the distance between the two objects is $0.1079$ $Units$