We know that acceleration of a falling body relative to earth is G*M_earth/(R_earth)^2 What would the acceleration formula if the earth's mass is same as sun's mass but with same earth's radius? Is there algabreac simple closed formula according to GR for the acceleration of falling body relative to an observer in rest relative to the earth?, or we just must solve GR differential equations numerically?
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As I understand from the other question you referred to is that the result is the acceleration of a held body at distance r relative to a free falling observer momentarily at rest relative to the held body. But I am asking the reverse question, what is the acceleration of the free falling abserver as measured by the non-inertial frame of the held body. Is it same result? – Ahmed Kamal Kassem May 30 '19 at 05:58
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Using the linked formula, in the Google calculator, I get a surface acceleration of 3264000 m/s^2. The GR correction, $1/\sqrt{1-r_s/r}$, is only 1.000232 – PM 2Ring May 30 '19 at 06:00
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Nice, I just need the symbolic formula for the acceleration of a free falling body with respect to the non-inertial earth (with sun mass) frame of reference. Is it the same linked formula? Cause the linked formula is the acceleration of earth's surface relative to the free falling observer – Ahmed Kamal Kassem May 30 '19 at 06:06
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Earth and sun are just example, my main interest is about the formula when the field is as strong as we can't neglect the GR correction – Ahmed Kamal Kassem May 30 '19 at 06:08
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Let us continue this discussion in chat. – PM 2Ring May 30 '19 at 06:10