Question

to 1 earth radius. The astronauts weight is An astronaut is floating happily outside her spaceship, which is orbiting the earth at a distance above the earths surface equal to 1 earth radius.
The astronauts weight is:

a.one-fourth her normal weight on earth

b.equal to her normal weight on earth

c.half her normal weight on earth

d.zero

Answer 1

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

$g = \dfrac{GM_{e}}{R^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

where 'G' is gravitational constant, '$M_{e}$' 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 '$g'$' be the value of the acceleration due to gravity at that height, then

$g' = \dfrac{G M_{e}}{(R + h)^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Taking the ratio of both the equations, and as given h = R.

$\dfrac{g'}{g} &=& \dfrac{g' = \dfrac{G M_{e}}{(R + h)^{2}}}{g' = \dfrac{G M_{e}}{(R + h)^{2}}}\\&=& \dfrac{R^{2}}{(R + h)^{2}}\\&=& R^{2}(R + R)^{2}\\&=& \dfrac{1}{4}$

So,

$&& g' = \dfrac{g}{4}\\&or,& mg' = \dfrac{mg}{4}$

where 'm' is the mass of the astronaut.

So the weight of the astronaut will be one-forth her normal weight on earth.