Naked Science Forum

Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: hamza on 30/05/2007 17:21:35

Title: A question - Earth's orbital period and distance
Post by: hamza on 30/05/2007 17:21:35

hi people.. i m new to this web site.. i found it quite interesting and informative.My question is that if the earth revolves arround the sun in an orbit with a radius half of the radius with which the earth is orbitting now... Than how many days will be their in one year 

Title: Re: A question - Earth's orbital period and distance
Post by: lightarrow on 30/05/2007 21:20:08

Quote from: hamza on 30/05/2007 17:21:35

hi people.. i m new to this web site.. i found it quite interesting and informative.My question is that if the earth revolves arround the sun in an orbit with a radius half of the radius with which the earth is orbitting now... Than how many days will be their in one year 

If I computed correctly, 129 days (365/√8).

P.S.
Welcome on this Forum hamza!

Title: Re: A question - Earth's orbital period and distance
Post by: Karen W. on 30/05/2007 22:00:47

Welcome Hamza And Good day to you Alberto!

Title: Re: A question - Earth's orbital period and distance
Post by: Seany on 31/05/2007 00:21:37

Welcome Hamza!! Great question [;D]

Title: Re: A question - Earth's orbital period and distance
Post by: syhprum on 31/05/2007 09:35:58

Please let me know how the factor 8^0.5 is derived in your equation, if I try to apply this to the orbital time of an Earth orbiting satellite compared to one in geostationary orbit I cannot make it work.

Failed O level maths 1944.

Title: A question - Earth's orbital period and distance
Post by: hamza on 31/05/2007 15:16:37

hey guys.. i found the answer.. infact light arrow is right.. the formula used is T(square) α R(cube).. where T is the number of days and R is the radius.. so we can take T(1) to be 365 days and T(2) is to be calculated. R(1) is the present radius of the orbit and R(2) is half of R(1) . Now divide both equations and u get 365/√8 ... 129 days.

Title: A question - Earth's orbital period and distance
Post by: syhprum on 31/05/2007 19:14:28

This does not compute for me!, if I take the orbit of a geosynchronous satellite (35,768km) and divide it by the height of a ground level satellite (6441 km)I get the ratio 5.553.
Cube this and the take the sqr gives me 13.086 but the ratio of orbital times is 1436/84 = 17.1 !! Where am I going wrong ?

Title: A question - Earth's orbital period and distance
Post by: syhprum on 31/05/2007 19:31:33

The penny has dropped. the distance of the geosynchronous satellite from the centre of the Earth should have been used not the height, when I use this the result comes out correct.
This calculation uses Keplers 3/2 power law