<p>I'm trying to solve this task:</p>
<blockquote>
<p>For amateur observations, a small telescope with a lens focal length of 120 cm was brought to the Byurakan Astrophysical Observatory. If this telescope is directed strictly to the zenith and the guidance is turned off, then the time during which the star crosses the field of view of the telescope is 2 min 38 s. Is it possible to observe the entire disk of the Moon with this telescope?</p>
</blockquote>
<p>I was thinking like that: <span class="math-container">$2 min$</span> <span class="math-container">$38 s = 158 s$</span>, one sidereal day is 23 hours 56 min 4 s (or 86164 s), so the star way is about 0.66 degrees (or 39') long. So it should be the field of view of the telescope. Does this mean that it is possible to observe Moon which diameter in the sky is maximum 32'? I am not sure as I didn't take in the account the focal length of the telescope that is in the task too, but I don't know where to use it. Could somebody help?</p>

CC BY-SA 4.0
2022-02-08T21:01:26.787Z
You need to know the latitude of the observatory for this. The circumference of a circle covered by a star decreases with cos(Latitude) so its speed on the sky decreases by the same amount. With a latitude of 45 deg for instance a star at zenith would only travel about 28' in 158 sec, not enough to fit in the moon.
Thomas
CC BY-SA 4.0
2022-02-08T21:04:32.263Z
You also need to consider how close the moon can get to the zenith, given the latitude of the observatory.
Dr Chuck
CC BY-SA 4.0
2022-02-08T21:11:17.850Z
@DrChuck I don't think the moon is supposed to be at the zenith as well. It is only about the field of view of the telescope here.
Thomas
CC BY-SA 4.0
2022-02-08T21:17:59.137Z
@DrChuck It is about the linear distance the star covers within a given time, not the angular distance (angular distance meaning the rotation angle with regard to the celestial pole, not the angular distance as seen by the observer)
Thomas
CC BY-SA 4.0
2022-02-09T07:48:44.207Z
@Thomas thanks.  I shouldn't try to do stuff in my head at night. Comment changed.
Dr Chuck
CC BY-SA 4.0
2022-02-15T17:03:13.800Z
Is this a trick question? The Moon doesn't come close to the zenith from that location.
Aaron F
<observational-astronomy><telescope>
Is it possible to observe the full Moon using this telescope?
ALiCe P.
# Is it possible to observe the full Moon using this telescope?

(Asked by: username_0 on 2022-02-08)

I'm trying to solve this task:

> For amateur observations, a small telescope with a lens focal length of 120 cm was brought to the Byurakan Astrophysical Observatory. If this telescope is directed strictly to the zenith and the guidance is turned off, then the time during which the star crosses the field of view of the telescope is 2 min 38 s. Is it possible to observe the entire disk of the Moon with this telescope?

I was thinking like that: $2 min$ $38 s = 158 s$, one sidereal day is 23 hours 56 min 4 s (or 86164 s), so the star way is about 0.66 degrees (or 39') long. So it should be the field of view of the telescope. Does this mean that it is possible to observe Moon which diameter in the sky is maximum 32'? I am not sure as I didn't take in the account the focal length of the telescope that is in the task too, but I don't know where to use it. Could somebody help?

**Tags:** observational-astronomy, telescope
> **Comments:**
> **username_1**: You need to know the latitude of the observatory for this. The circumference of a circle covered by a star decreases with cos(Latitude) so its speed on the sky decreases by the same amount. With a latitude of 45 deg for instance a star at zenith would only travel about 28' in 158 sec, not enough to fit in the moon.  
> **username_2**: You also need to consider how close the moon can get to the zenith, given the latitude of the observatory.  
> **username_1**: @DrChuck I don't think the moon is supposed to be at the zenith as well. It is only about the field of view of the telescope here.  
> **username_1**: @DrChuck It is about the linear distance the star covers within a given time, not the angular distance (angular distance meaning the rotation angle with regard to the celestial pole, not the angular distance as seen by the observer)  
> **username_2**: @Thomas thanks.  I shouldn't try to do stuff in my head at night. Comment changed.  
> **username_3**: Is this a trick question? The Moon doesn't come close to the zenith from that location.  

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## 0 answers

