Question

When the Moon is directly overhead at sunset, the force by the Earth on the Moon, FEM, is essentially 90° to the force by the sun on the moon, FSM, as depicted in the image. Let FEM = 1.58 × 1020 N, FSM = 4.56 × 1020 N, all other forces on the Moon be negligible, and the mass of the Moon be m = 7.35 × 1022 kg. (a) What is the magnitude of the net force exerted by the Earth and Sun on the Moon in Newtons? Part (b) What is the magnitude of the Moon's acceleration in m/s2?

Answer 1

given

angle = 90°

F_EM = 1.58 x 10²⁰ N

F_SM = 4.56 x 10²⁰ N

m = 7.35 x 10²² kg

according to Newton's second law

F_net = ma

F_net = ( (F_EM)² + (F_SM)² )^1/2

= ( (1.58 x 10²⁰ )² + ( 4.56 x 10²⁰ )² )^1/2

F_net = 4.825 x 10²⁰ N

so the magnitude of the net force exerted by

the Earth and Sun on the Moon in Newtons is F_net = 4.825 x 10²⁰ N

b )

F_net = ma

4.825 x 10²⁰ = 7.35 x 10²² x a

a = 4.825 x 10²⁰ / 7.35 x 10²²

a = 6.5646 x 10^-3 m /sec²

a = 0.0065646 m/sec²

the magnitude of the Moon's acceleration in m/sec² is a = 0.0065646 m/sec²

Answer 2

Given:

• Force by Earth on Moon ($F_{EM}$) = $1.58\times 10^{20}\text{N}$ (downward, toward Earth).

• Force by Sun on Moon ($F_{SM}$) = $4.56\times 10^{20}\text{N}$ (horizontal, toward Sun).

• Mass of Moon ($m$) = $7.35\times 10^{22}\text{kg}$.

• Angle between forces = $90^{\circ }$ (perpendicular).

• 
  

---

Part (a): Magnitude of Net Force

Since $F_{EM}$ and $F_{SM}$ are perpendicular, use the Pythagorean theorem:

$$F_{\text{net}}=\sqrt{F_{EM}^2+F_{SM}^2}$$$$F_{\text{net}}=\sqrt{(1.58\times 10^{20}{)}^2+(4.56\times 10^{20}{)}^2}$$$$F_{\text{net}}=\sqrt{2.496\times 10^{40}+20.79\times 10^{40}}$$$$F_{\text{net}}=\sqrt{23.29\times 10^{40}}=4.83\times 10^{20}\text{N}$$

---

Part (b): Moon's Acceleration

Newton’s second law:

$$a=\frac{F_{\text{net}}}{m}$$$$a=\frac{4.83\times 10^{20}}{7.35\times 10^{22}}=0.00657{\text{m/s}}^2$$

 

---

Final Answers:

(a) Net force = $4.83\times 10^{20}\text{N}$
(b) Moon’s acceleration = $6.57\times 10^{-3}{\text{m/s}}^2$