Question

The gravitational force that the sun exerts on the moon is perpendicular to the force that the earth exerts on the moon. The masses are: mass of sun=1.99 × 1030 kg, mass of earth=5.98 × 1024 kg, mass of moon=7.35 × 1022 kg. The distances shown in the drawing are rSM = 1.50 × 1011 m and rEM = 3.85 × 108 m. Determine the magnitude of the net gravitational force on the moon.

Answer 1

The gravitational force that the sun exerts on the moon is perpendicular to the force that the earth exerts on the moon. The masses are: mass of sun=1.99 × 1030 kg, mass of earth=5.98 × 1024 kg, mass of moon=7.35 × 1022 kg. The distances shown in the drawing are rSM = 1.50 × 1011 m and rEM = 3.85 × 108 m. Determine the magnitude of the net gravitational force on the moon.

Solution:

$$\begin{array}{l} \mathrm{M}_{\mathrm{s}} = 1.99 \times 10^{30} \mathrm{kg} - \text{mass of Sun}; \\ \mathrm{M}_{\mathrm{e}} = 5.98 \times 10^{24} \mathrm{kg} - \text{mass of Earth}; \\ \mathrm{M}_{\mathrm{m}} = 7.35 \times 10^{22} \mathrm{kg} - \text{mass of Moon}; \\ \mathrm{r}_{\mathrm{SM}} = 1.50 \times 10^{11} \mathrm{m} - \text{distance to the Moon from the Sun}; \\ \mathrm{r}_{\mathrm{EM}} = 3.85 \times 10^{9} \mathrm{m} - \text{distance to the Moon from the Earth}; \\ \mathrm{G} = 6.67 \times 10^{-11} \mathrm{N} \left(\frac{\mathrm{m}}{\mathrm{kg}}\right)^2 - \text{gravitational constant} \end{array}$$

The gravitational force that acts on the Moon by the Earth (Law of Gravity):

$$F_{\mathrm{e}} = G \frac{M_{\mathrm{e}} \cdot M_{\mathrm{m}}}{r_{\mathrm{SM}}^{2}} = 6.67 \times 10^{-11} \mathrm{N} \left(\frac{\mathrm{m}}{\mathrm{kg}}\right)^2 \cdot \frac{5.98 \times 10^{24} \mathrm{kg} \cdot 7.35 \times 10^{22} \mathrm{kg}}{(3.85 \times 10^{9} \mathrm{m})^2} = 1.98 \times 10^{20} \mathrm{N}$$

The gravitational force that acts on the Moon by the Sun (Law of Gravity):

$$F_{\mathrm{S}} = G \frac{M_{\mathrm{s}} \cdot M_{\mathrm{m}}}{r_{\mathrm{SM}}^{2}} = 6.67 \times 10^{-11} \mathrm{N} \left(\frac{\mathrm{m}}{\mathrm{kg}}\right)^2 \cdot \frac{1.99 \times 10^{30} \mathrm{kg} \cdot 7.35 \times 10^{22} \mathrm{kg}}{(1.50 \times 10^{11} \mathrm{m})^2} = 4.34 \times 10^{20} \mathrm{N}$$

Net gravitational force on the moon:

$$\vec{F} = \vec{F_{\mathrm{e}}} + \vec{F_{\mathrm{S}}}$$

Pythagorean theorem for a right triangle ABC:

$$F = \sqrt{F_{\mathrm{S}}^{2} + F_{\mathrm{e}}^{2}} = \sqrt{(1.98 \times 10^{20} \mathrm{N})^{2} + (4.34 \times 10^{20} \mathrm{N})^{2}} = 4.77 \times 10^{20} \mathrm{N}$$

Answer: magnitude of the net gravitational force on the moon is $4.77 \times 10^{20} \mathrm{N}$.