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Answer:
i am the guy from the time you asked the question for the first time. after discussing with my colleagues i finally came up with a answer so please give me a like
Explanation:
The answer is that it depends on what you assume happens to the Moon’s orbital period around the Earth.
The equations we use for satellites in basic school physics assume that the satellite mass is negligible compared with Earth. However this isn’t the case with the Moon, which is actually around 1.2% of Earth’s mass. So we need to use more accurate equations.
If we stick with the circular orbit approximation then the equation we need here is this one:
ω2R3=G(M+m)ω2R3=G(M+m) where ω is the orbital angular velocity, R in this case is the distance between Earth and Moon, M the Earth’s mass and mm the Moon’s mass.
The effect of doubling the Moon’s mass mm on this equation is that M+mM+m increases by around 1.2%. So this means that the left hand side of the equation must also increase by 1.2%.
If we take the case of the orbital period and therefore ωω stays the same, then it must be R3R3 that increases by 1.2%, which means that RR the distance increases by around 0.4%. That’s an increase on the nominal distance of 384400 km by around 1500 km.
At the other extreme we could assume that the distance stays the same in which case ω2ω2 must increase by 1.2%, which means that ωω increases by around 0.6%, or put another way the orbital period decreases by around 0.6%. The orbital period is the sidereal lunar month which is 27.3 days, and so this would decrease by around 0.16 days or around 4 hours.
i hope it helps ☺️☺️