Question

The distance between the International Space Station and the Earth surface is 356 km. The

mass and radius of the Earth is 5.97 x 1024 kg and 6.38 x 106 m respectively.

b)percentage difference between an astronaut’s weight on the Earth and on the space station if his mass is 70 kg,​

Answer 1

Answer:

The distance between the International Space Station and the Earth's surface is 356 km. The  mass and radius of the Earth are $5.9*10^{24}kg$ and $6.38*10^{6}m$ respectively.  The percentage difference between an astronaut’s weight on the Earth and on the space station if his mass is 70 kg, is $6.1$%

Explanation:

Acceleration due to gravity changes as we move from the earth's surface.

we have the formula to calculate the variation of acceleration due to gravity with altitude,

$g_{h}=g(\frac{R_{e} }{R_{e} +h } )$

where $g=9.8m/s^{2}$

The radius of the earth $R_{e} =6.38*10^{6}m$

The distance between the space station and the earth surface is $h=356km=0.35*10^{6} m$

substituting these values we get,$g_{h}=9.8(\frac{6.38*10^{6} }{6.38*10^{6}+0.35*10^{6} } ) =9.2m/s^{2}$

Now the weight of the astronaut at the earth's surface is $mass*g=70*9.8=686N$

weight of the astronaut at the space station = $70*g_{h}$$=70*9.2=644N$

The percentage difference between weight on the earth and on the space station is

$\frac{W_{e} -W_{h} }{W_{e} } *100$$\frac{686-644}{686} *100=6.1$%