ch13-p072 - M by 2 10 30 kg. Thus, admitting some tolerance...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
72. (a) The gravitational acceleration a g is defined in Eq. 13-11. The problem is concerned with the difference between a g evaluated at r = 50 R h and a g evaluated at r = 50 R h + h (where h is the estimate of your height). Assuming h is much smaller than 50 R h then we can approximate h as the dr which is present when we consider the differential of Eq. 13-11: | da g | = 2 GM r 3 dr 2 GM 50 3 R h 3 h = 2 GM 50 3 (2 GM/c 2 ) 3 h . If we approximate | da g | = 10 m/s 2 and h 1.5 m, we can solve this for M . Giving our results in terms of the Sun’s mass means dividing our result for
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: M by 2 10 30 kg. Thus, admitting some tolerance into our estimate of h we find the critical black hole mass should in the range of 105 to 125 solar masses. (b) Interestingly, this turns out to be lower limit (which will surprise many students) since the above expression shows | da g | is inversely proportional to M . It should perhaps be emphasized that a distance of 50 R h from a small black hole is much smaller than a distance of 50 R h from a large black hole....
View Full Document

Ask a homework question - tutors are online