# How to calculate when an object will be directly

• Wheelwalker
In summary: BzK8fSmzN4c7a1NzZEd3MmcjM/view?usp=sharingIn summary, the author is having trouble determining when a celestial object will be overhead (in terms of right ascension, not declination). He has found a workaround by converting longitude to hours and subtracting them from GMT to find local sidereal time. There are easier ways to do this, but he still needs to calculate LST when not near the equinox.
Wheelwalker
I'm having some trouble determining when a celestial object will be overhead (in terms of right ascension, not declination). Since we use solar time, it wouldn't be correct to simply use the right ascension of the object since each day the sidereal and solar time become four minutes different. So, given the celestial coordinates of an object, latitude and longitude of a location on earth, how do I find what time the object will be overhead? I've been grappling with this for awhile now and could really use some help instead of being asked more questions.

I've actually just made some progress on the problem from my book. I needed to determine the local sidereal time over Cuba at approximately 6 pm in late September. Using the GMT, and assuming (to make it easier) that this event happened on the equinox, I converted Cuba's longitude to hours and subtracted them from the GMT to find the approximate local sidereal time. Is there an easier way to do this? I feel like it would be an absolute pain to calculate LST when not near the equinox.

Wheelwalker said:
I've actually just made some progress on the problem from my book. I needed to determine the local sidereal time over Cuba at approximately 6 pm in late September. Using the GMT, and assuming (to make it easier) that this event happened on the equinox, I converted Cuba's longitude to hours and subtracted them from the GMT to find the approximate local sidereal time. Is there an easier way to do this? I feel like it would be an absolute pain to calculate LST when not near the equinox.

I wouldn't want to do it in my head, but here's a pretty simple formula to get the sidereal time at Greenwich for any date and time. From there, you just adjust for your longitude, as you did above.

If you don't understand it, just google the various terms such as "Julian date." You might also consider googling for "practical astronomy" or "ephemeris calculations" to get a more comprehensive tutorial on time conversion.

If you just want the answer without doing the calculation yourself, there are lots of sites that will do it for you, e.g.
http://www.jgiesen.de/astro/astroJS/siderealClock/

Yes you would need time but you would also need position. Azimuth and Altitude must be calculated from the Hour Angle, Latitude and Declination with Altitude to be 90 degrees.

HA = f(gst, time and Longitude minus the Right Ascension)
AZI = f(HA, Latitude, Declination)
ALT = f(HA, Latitude, Declination)

Last edited:
Wheelwalker said:
I've actually just made some progress on the problem from my book. I needed to determine the local sidereal time over Cuba at approximately 6 pm in late September. Using the GMT, and assuming (to make it easier) that this event happened on the equinox, I converted Cuba's longitude to hours and subtracted them from the GMT to find the approximate local sidereal time. Is there an easier way to do this? I feel like it would be an absolute pain to calculate LST when not near the equinox.

Local sidereal time (which is the same as right ascension) is the Greenwich Mean Sidereal Time plus the longitude.

You can look up the GMST at midnight for a particular date in an Astronomical Almanac, and then add the current Greenwich Mean Time (solar) to get your current GMST.

Or you can do the easy thing and download a spreadsheet that calculates GMST for any time you want at any given longitude just by setting the azimuth to 180. Or, if your real goal is correlate your location and your telescope's az/el, it will calculate the RA/dec of the object you're looking at directly instead of you having to figure it out manually. (You should do all of this manually at least once, just because, but I don't think I know anyone that would want to do this manually all the time.)

This has several useful spreadsheets. You want the ra_dec.xls spreadsheet to get right ascension & declination.

## 1. How do I calculate the distance an object will travel when thrown at a certain angle?

To calculate the distance an object will travel when thrown at a certain angle, you will need to use the formula: D = V₀² sin(2θ) / g, where D is the distance, V₀ is the initial velocity, θ is the angle of the throw, and g is the acceleration due to gravity (9.8 m/s²). This formula assumes ideal conditions and neglects air resistance.

## 2. How can I determine the time it will take for an object to reach a specific point?

The time it takes for an object to reach a specific point can be calculated using the formula: t = √(2d/g), where t is the time, d is the distance, and g is the acceleration due to gravity (9.8 m/s²). This formula also assumes ideal conditions and neglects air resistance.

## 3. What factors can affect the accuracy of my calculations for when an object will be directly?

The accuracy of your calculations can be affected by factors such as air resistance, which can slow down the object and change its trajectory. Other factors include the initial velocity, angle of throw, and any external forces acting on the object.

## 4. Can I use the same formula for all objects regardless of their mass?

Yes, the formula for calculating when an object will be directly can be used for all objects regardless of their mass. This is because the mass of an object does not affect its trajectory in ideal conditions.

## 5. Is there a difference in calculating when an object will be directly on Earth compared to on other planets?

Yes, there is a difference in calculating when an object will be directly on different planets. This is because the acceleration due to gravity varies from planet to planet. For example, on Mars, the acceleration due to gravity is 3.7 m/s², while on Earth it is 9.8 m/s². Therefore, the formulas for calculating distance and time will need to be adjusted accordingly.

• Astronomy and Astrophysics
Replies
4
Views
1K
• Astronomy and Astrophysics
Replies
12
Views
1K
• Astronomy and Astrophysics
Replies
56
Views
4K
• Astronomy and Astrophysics
Replies
3
Views
1K
• Astronomy and Astrophysics
Replies
142
Views
114K
• Astronomy and Astrophysics
Replies
2
Views
1K
• Astronomy and Astrophysics
Replies
7
Views
2K
• Astronomy and Astrophysics
Replies
6
Views
3K
• Astronomy and Astrophysics
Replies
2
Views
2K
• Astronomy and Astrophysics
Replies
17
Views
2K