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

Question #136276
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?
1
Expert's answer
2020-10-07T07:24:25-0400

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 xx , we apply x=r×cos(θ)x= r × cos(\theta)

=10.32×cos(0.32)= 10.32× cos(-0.32)


=9.7961=9.7961


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


=10.32×sin(0.32)= 10.32× sin(-0.32)


=3.2463= -3.2463


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


Now the first point p1=(9.7961,32463,45.3)p_1 = (9.7961, -32463, 45.3)


Second point;


Finding xx ; x=r×cos(θ)x= r × cos(\theta)


=10.33×cos(0.31)= 10.33× cos (-0.31)


=9.8376= 9.8376

`

Finding yy ; y=r×sin(θ)y = r× sin(\theta)


=10.33×sin(0.31)= 10.33 × sin (-0.31)


=3.1513=-3.1513


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


The second point, p2=(9.8376,3.1513,45.27)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;


D2=X2+Y2+Z2D^2 = X^2 + Y^2 + Z^2


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

X=9.83769.7961X= 9.8376-9.7961


=0.0415=0.0415


Y=3.15133.2463Y = -3.1513 - -3.2463


=0.095=0.095


Z=45.345.27Z = 45.3-45.27


=0.03=0.03


Now having the values, we calculate;


D2=X2+Y2+Z2D^2 = X^2 + Y^2 + Z^2


=0.04152+0.0952+0.032= 0.0415^2 + 0.095^2 +0.03^2


=0.00172225+0.009025+0.0009= 0.00172225 + 0.009025 +0.0009


=0.01164725= 0.01164725


D=0.01164725D = \sqrt{ 0.01164725}


=0.1079= 0.1079

Therefore the distance between the two objects is 0.10790.1079 UnitsUnits


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