the gravitational force between two asteroids is 6.2×10^8 n. asteroid y has three times the mass of asteroid…

Explanation:

Step1: Recall gravitational - force formula

The gravitational - force formula is $F = G\frac{m_1m_2}{r^2}$, where $F$ is the gravitational force, $G = 6.67\times10^{- 11}\ N\cdot m^2/kg^2$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between them. Let the mass of asteroid $Z$ be $m$, then the mass of asteroid $Y$ is $3m$. The distance $r = 2100\ km=2.1\times10^{6}\ m$ and $F = 6.2\times10^{8}\ N$.

Step2: Substitute values into the formula

Substitute into $F = G\frac{m_1m_2}{r^2}$: $6.2\times10^{8}=6.67\times10^{-11}\times\frac{m\times3m}{(2.1\times10^{6})^2}$. First, simplify the right - hand side: $6.2\times10^{8}=6.67\times10^{-11}\times\frac{3m^{2}}{4.41\times10^{12}}$. $6.2\times10^{8}=\frac{6.67\times3\times10^{-11}}{4.41\times10^{12}}m^{2}$. $\frac{6.67\times3\times10^{-11}}{4.41\times10^{12}}=\frac{20.01\times10^{-11}}{4.41\times10^{12}}\approx4.54\times10^{-23}$. So, $m^{2}=\frac{6.2\times10^{8}}{4.54\times10^{-23}}$. $m^{2}\approx1.37\times10^{31}$. $m=\sqrt{1.37\times10^{31}}\approx3.7\times10^{15}\ kg$.

Step3: Find the mass of asteroid $Y$

Since the mass of asteroid $Y$ is $3m$, then $m_Y = 3\times3.7\times10^{15}=1.1\times10^{16}\ kg$.

Answer:

$1.1\times10^{16}\ kg$