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If the moon starts revolving the earth with relativistic speed, does this mean that moon's gravitational field on earth will increase and maybe even overpower earth's own field due to increased mass from the effect of relativity

$$ m = \frac{m_0}{\sqrt(1 - \frac{v^2}{c^2})} $$

Ashutosh
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    Why would the Moon go that fast? A satellite can't get close enough to the centre of its primary to orbit at relativistic speed unless the primary is a neutron star or black hole. – PM 2Ring Dec 12 '23 at 10:35
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    related https://physics.stackexchange.com/questions/627548/can-relativistic-mass-actually-change-gravitational-pull – anna v Dec 12 '23 at 11:30
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    The downvoter possibly was motivated by the use of the obsolete concept of ‘relativistic mass’ and the impossibility for anything, let alone the Moon, to revolve (around) Earth at close to light speed. Instead if downvoting, comment. Apart from these issues this is a valid question. – my2cts Dec 12 '23 at 12:21

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Correct. The interaction depends on the speed of the moon in the suggested way for not too large Moon speeds.

Linearized gravity (Misner, Thorne, Wheeler, 'Gravitation', Ch 18.4) applies because the Earth has only weak gravity. Taking the Earth to be at rest, the interaction behaves like $$ \frac{GM}{r} \frac{m}{\sqrt{1-{v/c)^2}} \,.$$ With $v \approx 1 km/s$, the effect is only $1.1 \cdot 10^{-11}$. This approximation should hold up to quite high speed since the gravity of the Moon is so weak.

This follows from Eq. 18.15b in MTW. The second equality, '= Newton potential'), in this reference is only valid for low velocity.

G: gravitational constant

M: Earth mass

m: Moon mass

r: Earth - Moon distance

v: Moon velocity

my2cts
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