Question

4. An astronaut whose height h is 1.70 m floats feet "down" in an orbiting space shuttle at a distance center of Earth. 6.77 x 10° m from the (a) What is the dif her feet and at her head? ference in the gravitational acceleration at (b) If the astronaut is now feet "down" at the same orbital radius r of 6.77 x 10° m about a black hole of mass M1.99 x 10" kg (which is 10 times our Sun's mass), what is the difference in the gravitational acceleration at her feet and head? The black hole has a surface (called the horizon of the black hole) of radius R,2.95 x 10' m. Nothing, not even light, can escape from the surface or anywhere inside it. Note that the astronaut is (wisely) well outside the surface (at r 229R.)

Answer 1

The gravitational acceleration at the astronaut feet = g_f g_f = G M/r^2: M = earth mass = 5.972 e + 024 kg: G = 6.67408 times 10^-11 N m^2/kg^2 r = 6.77 e + 0.6 m implies g_f = 8.694879879 m/s^2 gravitational acceleration at astronaut feet = g_h g_h = G M/(r + h)^2 = 8.694875512 m/s^2: the difference (g_f - g_h) = 4.36684 e = 06 m/s^2 which is negligible b) if the astronaut is at orbit radius r = 6.77 e + 06 m and the mass of the black hole = M_b = 1.99 e + 0.31 kg the gravitational acceleration at the astronaut feet = g_f g_d f = G M_b/r^2: G = 6.67408 times 10^-11 N m^2/kg^2 Implies g_f = 28973226.66 e + 07 m/s^2 the gravitational acceleration at the astronaut head = g_h g_h = G M_b/(r + h)^2: h = 1.7 m G = 6.67408 times 10^-11 N m^2/kg^2 Implies g_h = 28973212.11 m/s^2 the difference is = g_h - g_f = 14.5548 m/s^2