Answer:
The astronaut's weight will be one-forth of her normal weight on earth.
Explanation:
From Newton's law of gravitation, we can write the acceleration due to gravity (g) on Earth's surface is given by
[tex]g = \dfrac{GM_{e}}{R^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)\\[/tex]
where 'G' is gravitational constant, '[tex]M_{e}[/tex]' is Earth's mass and 'R' is Earth's radius.
As shown in the figure, if the astronaut is at a height 'h' from earth's surface and if '[tex]g'[/tex]' be the value of the acceleration due to gravity at that height, then
[tex]g' = \dfrac{G M_{e}}{(R + h)^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Taking the ratio of both the equations, and as given h = R.
[tex]\dfrac{g'}{g} &=& \dfrac{g' = \dfrac{G M_{e}}{(R + h)^{2}}}{g' = \dfrac{G M_{e}}{(R + h)^{2}}}\\&=& \dfrac{R^{2}}{(R + h)^{2}}\\&=& R^{2}(R + R)^{2}\\&=& \dfrac{1}{4}\\[/tex]
So,
[tex]&& g' = \dfrac{g}{4}\\&or,& mg' = \dfrac{mg}{4}[/tex]
where 'm' is the mass of the astronaut.
So the weight of the astronaut will be one-forth her normal weight on earth.
