The radius Rhand mass Mh of a black hole are related by Rh = 2GMh/c^2, where c is the speed of light. Assume that the gravitational acceleration agof an object at a distance Ro = 1.001Rh from the center of a black hole is given by Ag = G M / r^2 (it is, for large black holes). (a) What is ag at ro for a very large black hole whose mass is 1.61 × 10^14 times the solar mass of 1.99 × 10^30 kg? (b) If an astronaut with a height of 1.66 m is at ro with her feet toward this black hole, what is the difference in gravitational acceleration between her head and her feet?(adsbygoogle = window.adsbygoogle || []).push({});

Rh = 2(6.67x10^-11)(3.204x10^44)/ (3.0x10^8)^2

Rh = 4.75 x 10^17

So then, Ag = GM/ (1.001Rh)^2

Ag = (6.67x10-11)(3.204x10^44)/(1.001 x 4.75x10^17)^2

Ag = 0.0945 m/s^2

I find myself having trouble with part B. How do i relate them?

I did Ag = (6.67x10^-11)(3.204x10^44)/(1.66 x 4.75x10^17)^2

Ag = 0.0344 m/s^2

So i did the difference btwn answer A and B and got .0601 m/s^2

But i know its wrong. Can someone help me out? Thanks!

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# Gravitation above Black Hole

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