Question

a spacecraft is sent to orbit around a planet that has a mass of 3.0 x 10^24 kg and a radius of 6.0 x 10^6 m. if the spacecraft orbits at a distance of 4.0 x 10^7 m from the center of the planet, what is the gravitational force acting on the spacecraft due to the planet? 2.8 x 10^7 n 1.3 x 10^12 n 6.7 x 10^-11 n 3.0 x 10^7 n

Explanation:

Step1: Recall gravitational - force formula

The gravitational - force formula is $F = G\frac{Mm}{r^{2}}$, where $G = 6.67\times10^{- 11}\ N\cdot m^{2}/kg^{2}$ (the gravitational constant), $M$ is the mass of the planet, $m$ is the mass of the spacecraft, and $r$ is the distance between the center of the planet and the spacecraft. Here, we assume the mass of the spacecraft is not given, but we can still calculate the force using the given values of $G$, $M$, and $r$. Given $M = 3.0\times10^{24}\ kg$, $r = 4.0\times10^{7}\ m$, and $G = 6.67\times10^{-11}\ N\cdot m^{2}/kg^{2}$.

Step2: Substitute values into the formula

$F=6.67\times 10^{-11}\frac{N\cdot m^{2}}{kg^{2}}\times\frac{3.0\times 10^{24}\ kg\times m}{(4.0\times 10^{7}\ m)^{2}}$. First, calculate $(4.0\times 10^{7}\ m)^{2}=16\times10^{14}\ m^{2}$. Then, $F = 6.67\times10^{-11}\times\frac{3.0\times 10^{24}}{16\times10^{14}}\ N$.

Step3: Simplify the expression

$6.67\times10^{-11}\times\frac{3.0\times 10^{24}}{16\times10^{14}}=\frac{6.67\times3.0}{16}\times10^{-11 + 24-14}\ N$. $6.67\times3.0 = 20.01$, and $\frac{20.01}{16}\approx1.25$. $-11 + 24-14=-1$. So $F\approx1.25\times10^{ - 1+10}\ N=1.25\times10^{10}\ N\approx1.3\times10^{10}\ N$.

Answer:

$1.3\times 10^{12}\ N$ (closest to the calculated value among the given options)