The radus R h and the mass M h of ablack hole are in acordance withR h =2GM h /c

Question

The radus Rh and the mass Mh of ablack hole are in acordance withRh=2GMh/c2, where c is thevolocity of the light. Suppose that the gravitacional acelerationag of an object at a distancer0=1.001Rh, from the center of the blackhole, is given by the ecuationag=GM/r2 (for big black holes).
a) Find an expresion for ag inr0 in terms Mh. b) Increase or decrease ag in r0,with an increase of Mh? c) Wich is ag in r0, for a big blackhole hose mass is 1.55x1012 times the solar mass1.99x1030kg? d) If an astronaut is 1.70 m is in r0, withhis feet pointing to that black hole, wich is the gravitationalaceleration diference between his head and his feet? e) Is it seriously the tendence of the astronaut tostrecht?
a) Find an expresion for ag inr0 in terms Mh. b) Increase or decrease ag in r0,with an increase of Mh? c) Wich is ag in r0, for a big blackhole hose mass is 1.55x1012 times the solar mass1.99x1030kg? d) If an astronaut is 1.70 m is in r0, withhis feet pointing to that black hole, wich is the gravitationalaceleration diference between his head and his feet? e) Is it seriously the tendence of the astronaut tostrecht?

Explanation / Answer

a) r0 = 1.001Rh =2.002GMh/c2     ag =GMh/r02 =GMh/(2.002GMh/c2)2 =c4/(4.008004 GMh) b) As Mh increases, the denominator in the equationabove increases, so agdecreases. c) ag =(3×108)4/(4.008004(6.67×10-11)(1.55×1012 ×1.99×1030)) ˜ 9.82 m/s2 d) location of head = r0 + (1/2)(1.70) =2.002GMh/c2 + 0.85                              =2.002(6.67×10-11)(1.55×1012 ×1.99×1030)/(3×108)2 +0.85 ˜ 4.58×1015 m     location of feet = r0 -(1/2)(1.70) = 2.002GMh/c2 - 0.85                            =2.002(6.67×10-11)(1.55×1012 ×1.99×1030)/(3×108)2 -0.85 ˜ 4.58×1015 m     There is so little difference in distancethat there is no appreciable difference in acceleration due togravity. e) No. (Large objects such as stars orbiting the black hole dostretch, however.) d) location of head = r0 + (1/2)(1.70) =2.002GMh/c2 + 0.85                              =2.002(6.67×10-11)(1.55×1012 ×1.99×1030)/(3×108)2 +0.85 ˜ 4.58×1015 m     location of feet = r0 -(1/2)(1.70) = 2.002GMh/c2 - 0.85                            =2.002(6.67×10-11)(1.55×1012 ×1.99×1030)/(3×108)2 -0.85 ˜ 4.58×1015 m     There is so little difference in distancethat there is no appreciable difference in acceleration due togravity. e) No. (Large objects such as stars orbiting the black hole dostretch, however.)

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