Answer:
Magnitude of gravitational force between this planet and its moon: approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].
Acceleration of this moon towards the planet: approximately [tex]5.49 \times 10^{-3}\; \rm m \cdot s^{-2}[/tex].
Acceleration of this planet towards its moon: approximately [tex]2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].
Explanation:
Look up the gravitational constant, [tex]G[/tex]:
[tex]G \approx 6.67 \times 10^{-11}\; \rm m^3\cdot kg^{-1} \cdot s^{-2}[/tex].
Assume that both this planet and its moon are spheres of uniform density. When studying the gravitational interaction between this planet and its moon, this assumption allows them to be considered as two point masses.
The formula for the size of gravitational force between two point masses [tex]m_1[/tex] and [tex]m_2[/tex] with a distance of [tex]r[/tex] in between is:
[tex]\displaystyle F = \frac{G \cdot m_1 \cdot m_2}{r^2}[/tex],
where [tex]G[/tex] is the gravitational constant.
Let [tex]m_1[/tex] and [tex]m_2[/tex] denote the mass of this planet and its moon, respectively.
Calculate the size of gravitational force between this planet and its moon:
[tex]\begin{aligned} F &= \frac{G \cdot m_1 \cdot m_2}{r^2} \\ &\approx \frac{6.67 \times 10^{-11}\; \rm m^3 \cdot kg^{-1} \cdot s^{-2} \times 6.00 \times 10^{24}\; \rm kg \times 2.50 \times 10^{22}\; \rm kg}{{\left(2.70 \times 10^{8}\; \rm m\right)}^2} \\ &\approx 1.37 \times 10^{20}\; \rm N\end{aligned}[/tex].
Assume that other than the gravitational force between this planet and its moon, all other forces (e.g., gravitational force between this planet and the star) are negligible. The magnitude of the net force on the planet and on the moon should both be approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].
Apply Newton's Second Law of motion to find the acceleration of this planet and its moon:
[tex]\displaystyle \text{acceleration} = \frac{\text{net force}}{\text{mass}}[/tex].
For this moon:
[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{2.50 \times 10^{22}\; \rm kg} \approx 5.49\times 10^{-3}\; \rm m \cdot s^{-2}[/tex].
For this planet:
[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{6.00 \times 10^{24}\; \rm kg} \approx 2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].
