Question

homework 3 begin date: 8/17/2025 11:59:00 pm due date: 9/9/2025 11:59:00 pm end date: 9/9/2025 11:59:00 pm
problem 3: (6% of assignment value)
when the moon is directly overhead at sunset, the gravitational force on the moon due to the earth, $\vec{f}_{\text{em}}$, is perpendicular to the gravitational force on the moon due to the sun, $\vec{f}_{\text{sm}}$. these two forces have magnitudes $f_{\text{em}} = 1.98 \times 10^{20}$ n and $f_{\text{sm}} = 4.36 \times 10^{20}$ n. assume that all other forces are negligible. the mass of the moon is $m = 7.35 \times 10^{22}$ kg.

part (a)
write an expression for the magnitude of the acceleration of the moon.
a = input box
diagram of moon with $\vec{f}_{\text{em}}$ arrow
hints: 4% deduction per hint. hints remaining: 1
feedback: 1 for a 5% deduction
given the numerator $(2 f_{\text{em}} + 2 f_{\text{sm}})^{0.5}$:

• it appears that this numerator involves other terms. please think about what else should be included in the numerator

Explanation:

Step1: Recall Newton's Second Law

Newton's second law states that the net force \(\vec{F}_{\text{net}}\) acting on an object is equal to the mass \(m\) of the object multiplied by its acceleration \(\vec{a}\), i.e., \(\vec{F}_{\text{net}} = m\vec{a}\). The magnitude of the acceleration \(a\) is then \(a=\frac{F_{\text{net}}}{m}\).

Step2: Find the magnitude of the net force

The two forces \(\vec{F}_{\text{EM}}\) and \(\vec{F}_{\text{SM}}\) are perpendicular to each other. To find the magnitude of the net force, we use the Pythagorean theorem. If two vectors with magnitudes \(F_1\) and \(F_2\) are perpendicular, the magnitude of their resultant (net force) is \(F_{\text{net}}=\sqrt{F_1^2 + F_2^2}\). Here, \(F_1 = F_{\text{EM}}\) and \(F_2 = F_{\text{SM}}\), so \(F_{\text{net}}=\sqrt{F_{\text{EM}}^2+F_{\text{SM}}^2}\).

Step3: Substitute net force into acceleration formula

Substitute \(F_{\text{net}}=\sqrt{F_{\text{EM}}^2 + F_{\text{SM}}^2}\) into \(a = \frac{F_{\text{net}}}{m}\). We get \(a=\frac{\sqrt{F_{\text{EM}}^2 + F_{\text{SM}}^2}}{m}\).

Answer:

\(\frac{\sqrt{F_{\text{EM}}^2 + F_{\text{SM}}^2}}{m}\)