The mass of Earth is $m_{E} = 5.97 times 10^{24} mathrm{~kg}$.The radius of Earth is $R_{E} = 6.37 times 10^{6} mathrm{~m}$. The height of the space station is $h = 370 mathrm{~km}$.

$textbf{Required: }$

Finding the acceleration due to the gravity at the altitude of the International Space Station’s orbit.

According to Newton’s second law, the force due to the gravity (Weight) is given by

$$
begin{align*}
F &= W \
&= m_{1} ~ a \
end{align*}
$$

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{E} }{ r^{2} } \
&= m_{1} ~ a \
&= G ~ dfrac{ m_{1} ~ m_{E} }{ left( R_{E} + h right)^{2} } \
end{align*}
$$

Rearrange and solve for the acceleration due to the gravity at the altitude of the International Space Station’s orbit:

$$
begin{align*}
a &= G ~ dfrac{ m_{E} }{ left( R_{E} + h right)^{2} } \
&= 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} ~ dfrac{ 5.97 times 10^{24} mathrm{~kg} }{ left( 6.37 times 10^{6} mathrm{~m} + 3.7 times 10^{5} mathrm{~m} right)^{2} } \
&= 8.766 mathrm{~m/s^{2}}
end{align*}
$$

So, the acceleration due to the gravity at the altitude of the International Space Station’s orbit is $8.766 mathrm{~m/s^{2}}$.

The mass of Earth is $m_{E} = 5.97 times 10^{24} mathrm{~kg}$.The radius of Earth is $R_{E} = 6.37 times 10^{6} mathrm{~m}$. The acceleration due to gravity is $a = dfrac{g}{2}$.

$textbf{Required: }$

Finding the altitude above Earth’s surface (height).

According to Newton’s second law, the force due to the gravity (Weight) is given by

$$
begin{align*}
F &= W \
&= m_{1} ~ a \
end{align*}
$$

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{E} }{ r^{2} } \
&= m_{1} ~ a \
&= G ~ dfrac{ m_{1} ~ m_{E} }{ left( R_{E} + h right)^{2} } \
end{align*}
$$

Rearrange and solve for the acceleration due to the gravity at the altitude of the International Space Station’s orbit:

$$
begin{align*}
a &= G ~ dfrac{ m_{E} }{ left( R_{E} + h right)^{2} } \
&= dfrac{g}{2} \
end{align*}
$$

Rearrange and solve for the altitude above Earth’s surface (height):

$$
begin{align*}
left( R_{E} + h right)^{2} &= dfrac{ 2 ~ G ~ m_{E} }{g} \
h &= sqrt{ dfrac{ 2 ~ G ~ m_{E} }{g}} – R_{E} \
&= sqrt{ dfrac{ 2 times 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} times 5.97 times 10^{24} mathrm{~kg} }{9.8 mathrm{~m/s^{2}} }} – 6.37 times 10^{6} mathrm{~m} \
&= 2.645 times 10^{6} mathrm{~m}
end{align*}
$$

So, the altitude above Earth’s surface (height) is $2.645 times 10^{6} mathrm{~m}$.