Question

1. [5 marks] Astronomers have estimated that the nucleus of comet Hale-Bopp had a mass of 2.10 x 1015 kg and that it lost mass at an average rate of about 350,000 kg/s during the 100 days that it spent on its most recent pass close to the Sun. Estimate the total amount of mass lost during this encounter with the sun. If it loses this amount of mass with each encounter with the sun, how many orbits can it make before it completely disintegrates into teeny tiny bits?

Answer 1

Mass of the comet nucleus = 2.10 x 10¹⁵ kg

Average rate of mass loss = 350,000 kg/s = 3.5 x 10⁵ kg/s

Number of seconds in 100 days = 100 x 24 x 60 x 60 = 864,0000 s = 8.64 x 10⁶ s

We have converted 100 days to seconds because the mass loss rate of the comet is in seconds.

Therefore, total mass lost in 100 days = 3.5 x 10⁵ kg/s x 8.64 x 10⁶ s = 30.24 x 10⁶ x 10⁵ kg

   = 30.24 x 10¹¹ kg

Therefore the possible no of orbits = Total mass of the comet nucleus / mass lost in one orbit

=2.10 x 10¹⁵ / 30.24 x 10¹¹

= 6.9444 x 10²

= 694.44

Therefore it can complete 694.44 orbit before it completely disintegrates.