Question

the international space station has a mass of 419,455 kg. when it orbits the earth at an altitude of 400,000 m, what is the approximate gravitational force on the station due to earths gravity? (recall that earth has a radius of 6.37×10^6 m, it has a mass of 5.97×10^24 kg, and g = 6.67×10^(-11) n·m²/kg².)

a. 2.0×10^6 n
b. 3.6×10^6 n
c. 3.9×10^6 n
d. 2.8×10^6 n

Explanation:

Step1: Calculate the distance from the center of the Earth to the station

The distance $r$ is the sum of the Earth's radius $R_E$ and the altitude $h$. So $r=R_E + h$, where $R_E = 6.37\times10^{6}\text{ m}$ and $h = 400000\text{ m}=4\times 10^{5}\text{ m}$. Then $r=(6.37\times 10^{6}+4\times 10^{5})\text{ m}= 6.77\times 10^{6}\text{ m}$.

Step2: Use the law of universal gravitation

The law of universal gravitation is $F = G\frac{Mm}{r^{2}}$, where $G = 6.67\times 10^{-11}\text{ N}\cdot\text{m}^{2}/\text{kg}^{2}$, $M = 5.97\times 10^{24}\text{ kg}$ (mass of the Earth), $m = 419455\text{ kg}$ (mass of the space - station), and $r = 6.77\times 10^{6}\text{ m}$.
Substitute the values into the formula:
\[

$$\begin{align*} F&=6.67\times 10^{-11}\times\frac{5.97\times 10^{24}\times419455}{(6.77\times 10^{6})^{2}}\\ &=6.67\times 10^{-11}\times\frac{5.97\times 10^{24}\times419455}{4.58329\times 10^{13}}\\ &=6.67\times 10^{-11}\times\frac{5.97\times419455\times 10^{24}}{4.58329\times 10^{13}}\\ &=6.67\times 10^{-11}\times\frac{2.50414635\times 10^{30}}{4.58329\times 10^{13}}\\ &=6.67\times 10^{-11}\times5.46363\times 10^{16}\\ &=(6.67\times5.46363)\times10^{-11 + 16}\\ &=36.4424\times10^{5}\\ &\approx3.6\times 10^{6}\text{ N} \end{align*}$$

\]

Answer:

B. $3.6\times 10^{6}\text{ N}$