Astronomy

Detecting an asteroid

Detecting an asteroid
There were lots of collisions happened due to the asteroids coming from the outer space. Some incidents are very harmful for life on Earth. Now I try to investigate a new method to detect such asteroids coming from outer space and measure the velocity it moves.
We know there is relative motion of Earth with respect to the asteroid; due to Earth’s spinning and due to the revolution of Earth around sun. We know when an asteroid is coming towards the Earth, some energized electromagnetic waves should come towards the Earth. And we detect and measure those electromagnetic waves at two different Earth times. Then within that two different Earth times, Earth has gone some length along the orbit of the Earth as well as it has rotated some angle around the rotating axis of the Earth. But basically the asteroid moves in a straight path with respect to other stellar object. i.e. relative to sun. The situation is as below.
Picture

Where A0 is the position that initial observation of asteroid and A is the 2nd moment of asteroid detection and A and A0 same position on Earth but at two different Earth times. And within that time interval Earth has rotated ß angle around its axis of spinning. B is the position such that located left with respect to the rotating axis. And ? is the angle between A0O and asteroid coming direction. And ? is the angle between A0O and tangent line to Earth at A0(A0Y is the tangential line at point A0 to the Earth and OY is the perpendicular line to A0Y). And AX is the tangential line to Earth at point A and O’X is the perpendicular line for line AX.

Earth spins about 360 degrees within 24 hours. Therefore

 

ß = [2p/24) -/+ 2p/(365.25*24) ] * t ;   t is the Earth time difference of detecting asteroid at two different moments. And (-/+) depends on the considering direction of revolving Earth around sun.



Let W is the tangential velocity of Earth at point A
0 , W’ is the tangential velocity of Earth at point ‘A’ within space time. And U is the velocity of the asteroid with respect to ‘Earth detection’ at the first moment of detection at point A0. i.e. U was the velocity of the asteroid whenever now coming EM waves toward point A0 were emitted. And U’ is similar to the 2nd detection of EM waves at point A. And important : U is the velocity component of V at A0 in the direction A0Y and U’ is also the velocity component of V, at position A, in the direction A0Y.



Then U = V cos(?).cos(?)………………………(1)




U’ = – V’.cosd.cos a.cos ß. (minus include : because asteroid velocity component at the 2
nd detection is in the opposite direction of A0Y)



V’ is the velocity of the asteroid with respect to the Earth ,whenever now coming(at the 2
nd detection of EM waves coming from the asteroid) EM waves were emitted. We can consider



V’ =k.V k is a complex constant or function of V. Therefore




U’ = -kV.cosd.cos a.cos ß……………………(2)




ß explanation :

Picture

When Earth rotate ß angle around its axis of rotation the tangential line at point A0 should also rotate an angle ß. Therefore angle between A0Y and AX is also ß. Then the velocity component in the direction AX is equivalent to cos ß times that velocity component and that is in the direction A0Y. When time value ‘t’ is very small, we can neglect the second term of the expression for ß. Because when t is very small, the angle Earth has gone around sun is negligible compare to the angle Earth has rotated around its axis within time t.

We know Earth has two velocity components.

        1. Due to axial rotation there is a tangential velocity component at point A0.
        2. Due to the revolution around the sun , there is an another velocity component.

Revolution component of velocity W (along the direction A
0Y ) = r0*(2p/365.25*24)* cos ?

Where ? = (2p*t
1)/24

Explanation :
Picture

We know definitely PP0 and XX0 can be consider as perpendicular lines. Then we know that angle ? = (angular velocity * time taken to rotate the Earth about an angle ?). r0 is the distance between sun and Earth when 2nd detection occurs.

Angular velocity of Earth to spin about its axis = 2p/T


time taken to rotate the Earth about an angle ? = t
1 = time difference between the moment of asteroid 1st detection occurs and the time of the place such that the sun is overhead (whenever 1st detection is occurring) That (t1) can be easily measure: just the time difference between two clocks placed at A0 and sun overhead position.

Similarly revolution component of velocity W’ along the direction A
0Y =
r
0’ * (2p/365.25*24)*cos ?’ *cos ß

Where r
0’ is the distance between sun and Earth whenever the 2nd detection occurs.
And ?’ = (2p*t2/24)


Where t
2 is the time difference between the position of 2nd detection occurs and similar sun overhead position.

Tangential velocity component of W = R(2p/24)
Tangential velocity component of W’ = R(2p/24).cos ß
R is the mean radius of the Earth.
Then,


W = [ R(2p/24) + r
0*(2p/365.25*24)* cos ? ]…………………(3)
W’ = [ R(2p/24).cos ß + r
0’ * (2p/365.25*24)*cos ?’ *cos ß]………………………(4)

We can measure the values of ? and a : By measuring the angle between the flat Earth surface at point A
0 and the direction of asteroid observable; we can measure the angle ?. And by measuring the angle between the flat Earth surface at point A and the direction of asteroid observable ; we can directly measure the angle a.

We can measure the intensity of coming EM waves towards the point A
0(I1) by using suitable instrument as well as intensities of EM waves coming towards point A( I2 ) We have,
When ? ?0, I1 ? I
1,max and when d ?0, I2 ?I2, max. Therefore using the equation of maximum intensity and the intensity equations for point A0 and A, we can calculate the values of ? and d. Then we can get expressions for U and U’ in terms of V. We can directly get the values of Wand W’ as constant values. Let W= k1 and W’ = k2. Then the relative velocity of Earth with respect to the asteroid at position A0= V1 = k1 – g1(V)…………………………(5)

Then the relative velocity of Earth with respect to the asteroid at position A =


V
2 = k2– g2(V)…………………………(6)Where g

1(V) is the expression get from equation (1) and g2 (V) is the expression get from the equation (2). Then by relativistic Doppler formula we get

f
1 = [ 1/ v(1- (V1/C) ) ]. { 1- (W/U) }. f0
f2 = [ 1/ v(1- (V2/C) ) ]. { 1- (W’/U’) }. f0

Where f
1 is the frequency of coming EM waves toward point A0 detected at point A0. And f2 is similar frequency at point A. And f0 is the real frequency of EM waves coming towards A0, when they were emitting. And f0’ is the real frequency of the EM waves coming towards A, when they were emitting by asteroid. Since time difference between two detection of asteroid is very small , we can consider f0 = f0’ .Therefore by the above equations,
f
1 / f2 = [ 1/ v(1- (V1/C) ) ]. { 1- (W/U) } / [ 1/ v(1- (V2/C) ) ]. { 1- (W’/U’) }



Therefore we get the relation that contains V and k(V)terms. Therefore if we assume 

k(V) =1( that means velocity of the asteroid does not change within that time interval of two detections occurs ), then we can estimate the value of V.

By substituting that V value to equation (1) we can find value of f
1. And after that if we substitute that V value, to equation (2) and we can find value of f2. Then we get the difference of f1 and f2. If f1< f2 ; then we can conclude that that asteroid gradually becomes energize(may not be happened) , if f2< f1 ; then we can conclude that the asteroid gradually becomes weaker in energy.

If we assume f
1 = f2 directly, then by substituting that V value to both equations and by comparing those k(V) value with the integer orders of V value we got, we can get rough idea about the function k(V). Then by using some equations in relative motions and etc. we can estimate how far the asteroid from the Earth roughly.

Geerasee Wijesooriya
Geerasee is a guest contributor at Nerdynaut.
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11 Comments

11 Comments

  1. Ian

    June 4, 2015 at 2:37 am

    wow a rich article

  2. K.H.K. Geerasee Wijesuriya

    June 4, 2015 at 11:17 am

    Thank you Ian !

  3. Eranga Thilina

    June 4, 2015 at 4:54 pm

    There could be no errors with this method. Method is theoretical. But as astronomers we use more photometric investigations to detect asteroids and potentially hazardous objects (PHO) instead of motion calculations which is based on velocity and gravity because sudden gravitational variations can be happen on these kind of objects due to external objects. So calculated values can be underestimated or overestimated. Normally we calculate orbits and other properties of these kind of objects (specially for PHO) in to highly accurate manner. So we give high priority for Photometric investigations. There are so many software are designed and developed specially for Asteroid detection using raw data image files. Anyway appreciate this method. Best regards ~ Eranga Thilina ( Astrophysicist)

    • K.H.K. Geerasee Wijesuriya

      June 5, 2015 at 1:10 am

      Thank you for your comment and suggestions!

  4. Rex De Silva

    June 17, 2015 at 1:32 am

    Very interesting. Thanks for the maths. Do you know that many asteroids can be viewed easily with simple equipment like binoculars. Over the years I have observed on numerous occasions asteroids Vesta, Ceres, Pallas and on one occasion asteroid Irene.

    • K.H.K. Geerasee Wijesuriya

      June 17, 2015 at 6:16 am

      Thank you very much for your comment Mr/Mrs Rex. Yes there are several equipments those can be used to observe asteroids. Those have different capabilities to observe them with different resolution powers. Thank you again !

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