Question

question 20 of 20
a deep - space satellite is sent to orbit a distant planet with unknown mass. on arrival, the satellite begins its orbit and measures a gravitational pull from the planet of 620 n. if the satellite has a mass of 450 kg and orbits the planet at a radius of 4.0×10^6 m, what is the approximate mass of the planet? (recall that g = 6.67×10^(-11) n·m2/kg2.)
a. 7.8×10^23 kg
b. 3.3×10^23 kg
c. 2.9×10^22 kg
d. 4.5×10^24 kg

Explanation:

Step1: Recall gravitational force formula

The gravitational force formula is $F = G\frac{Mm}{r^{2}}$, where $F$ is the gravitational force, $G$ is the gravitational - constant, $M$ is the mass of the planet, $m$ is the mass of the satellite, and $r$ is the radius of the orbit. We need to solve for $M$.

Step2: Rearrange the formula for $M$

Rearranging the formula $F = G\frac{Mm}{r^{2}}$ for $M$ gives $M=\frac{F\times r^{2}}{G\times m}$.

Step3: Substitute the given values

We are given that $F = 620\ N$, $G=6.67\times 10^{-11}\ N\cdot m^{2}/kg^{2}$, $m = 450\ kg$, and $r = 4.0\times 10^{6}\ m$.
Substitute these values into the formula:
\[

$$\begin{align*} M&=\frac{620\times(4.0\times 10^{6})^{2}}{6.67\times 10^{-11}\times450}\\ &=\frac{620\times16\times 10^{12}}{6.67\times 10^{-11}\times450}\\ &=\frac{9920\times 10^{12}}{3001.5\times 10^{-11}}\\ &=\frac{9.92\times 10^{15}}{3.0015\times 10^{-11}}\\ &=\frac{9.92}{3.0015}\times10^{15 + 11}\\ &\approx3.3\times 10^{23}\ kg \end{align*}$$

\]

Answer:

B. $3.3\times 10^{23}\ kg$