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If I knew this, I wouldn't ask question, because I would have enougb factors in the equation, to solve. If I was a physicists.

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Note that the mass $m$ of an object is invariant to its w.r.t. its position relative to Earth.

But weight $W$, the attractive force exerted on the object by Earth, is not:

$$W(h)=\frac{GM_e}{(R_e+h)^2}m$$

is the force exerted by Earth on the object, $h$ $\text{meter}$ above the Earth's surface.

For $h=0$ we get: $$W(0)=\frac{GM_e}{R_e^2}m$$

Clearly for $h > 0$, then $W(h) < W(0)$.

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  • $\begingroup$ At that elevation (ignoring rotation), weight is reduced by ~0.2%, according to Google $\endgroup$
    – PM 2Ring
    Jan 24, 2023 at 19:12
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No. You are further away from Earth, so the gravitational pull is less. While your mass is the same, the weight force you feel depends on where you are.

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