Question

If the 78.0 kg astronaut were in a spacecraft 6R from the center of the earth, what would the astronaut's weight be on earth? 764.4 N760 N764 N in the spacecraft? 21.24 N21 N21.3 N

Answer 1

(a) 764.4 N

The weight of the astronaut on Earth is given by:

$F=mg$

where

m is the astronaut's mass

g is the acceleration due to gravity

Here we have

m = 78.0 kg

g = 9.8 m/s^2 at the Earth's surface

So the weight of the astronaut is

$F=(78.0)(9.8)=764.4 N$

(b) 21.1 N

The spacecraft is located at a distance of

$r=6R$

from the center of Earth.

The acceleration due to gravity at a generic distance r from the Earth's center is

$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass.

We know that at a distance of r = R (at the Earth's surface) the value of g is 9.8 m/s^2, so we can write:

$GM=9.8R^2$ (1)

the acceleration due to gravity at r=6R instead will be

$g'=\frac{GM}{(6R)^2}$

And substituting (1) into this formula,

$g'=\frac{9.8R^2}{36R^2}=0.27 m/s^2$

So the weight of the astronaut at the spacecratf location is

$F'=mg'=(78.0 kg)(0.27 m/s^2)=21.1 N$