NGC 4546 [R.A. = 12h35m29.5s, Dec. = -03d47m35s] has a very compact (point sourc

Question

NGC 4546 [R.A. = 12h35m29.5s, Dec. = -03d47m35s] has a very compact (point source like from the ground) companion NGC 4546-UCD1 [R.A. = 12h35m2S.7s, Dec. = -03d47m21.Is] which is now known to be the remnant centre of a once much more massive galaxy, which was tidally destroyed by the gravity of NGC 4546. Calculate the zenith angle and airmass of NGC 4546 when observed at an 1ST of 0 hrs (i.e. at upper culmination, when it is at its highest in the sky) from the Large Binocular Telescope [latitude: 32.7 degree N]. Astronomers prefer to observe astronomical objects at airmass as close to 1 as possible. The most accurate observations are often limited to airmass = 1.41 (zenith angle = 45 degree). For how long is NGC 4546 above airmass 1.41? Other than atmospheric refraction, name and describe two other effects which are worsened by observing at large zenith angles.

Explanation / Answer

a) we know that zenith angle Z is given by the expression

secZ=1/[SinSin+CosCosCosh]

here = lattitude of observation= 32.7degreeN

= declination of the object=-3d47m35s=-262055s= -72.79 degree

h= hour angle of the object=sidereal time- right ascension= 0hrs- 12h35min29.5sec =12h35min29.5sec = 12.59degree

using all values we get secZ =1/[Sin32.7Sin(-72.79)+Cos32.7Cos(-72.79)Cos112.59]

or cosZ=[Sin32.7Sin(-72.79)+Cos32.7Cos(-72.79)Cos12.59]= -.277

or Z=arccos[-.277]= 106.08 degree

Again air mass=1/Cos zenith angle =-.277

b) It wwill be above airmass1.41 for 2.35 hrs

c) At large zenith angle surface insolation and radio wave activities are worsened.

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