Question

An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above the surface of Earth?

Answer 1

Answer:

$mg=200.4 N$.

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$, $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$,

$mg=8*10^{2}$,

$m=(8*10^{2})/g$,

since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$.

Now we can use Newton's law of universal gravitation

$F=G\frac{Mm}{r^{2}}$,  

where $m$ is the mass of the astronaut and $M$ is the mass of the earth. From Newton's second law we know that

$F=ma$,

in this case the acceleration is the gravity so

$F=mg$, (becarefull, gravity at this point is no longer $9.8m/s^{2}$ because we are not in the surface anymore)

and this get us to

$mg=G\frac{Mm}{r^{2}}$, where $mg$ is his new weight.

We need to remember that the mass of the earth is $M=5.972*10^{24}kg$ and its radius is $6.37*10^{6}m$.

The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$,

$mg=200.4 N$.