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, $\vec{f}_{em}$, is perpendicular to the gravitational force on the moon due to the sun, $\vec{f}_{sm}$. these two forces have magnitudes $f_{em} = 1.98 \times 10^{20}$ n and $f_{sm} = 4.36 \times 10^{20}$ 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.
$a = \sqrt{(2 \cdot f_{em} + 2 \cdot f_{sm})^{0.5}}$ $\boldsymbol{\times}$ incorrect!

• part (b)

what is the acceleration of the moon in m/s²?
$a = \square$

(diagram: moon with $\vec{f}_{em}$ (horizontal right) and $\vec{f}_{sm}$ (vertical down))

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Explanation:

Step1: Find the net force

The two forces \( F_{EM} \) and \( F_{SM} \) are perpendicular, so the magnitude of the net force \( F_{net} \) is given by the Pythagorean theorem. But wait, the problem statement in part (a) had a wrong expression, let's correct it. The correct net force magnitude when two forces \( F_1 \) and \( F_2 \) are perpendicular is \( F_{net}=\sqrt{F_{EM}^2 + F_{SM}^2} \)? Wait no, wait the problem's part (b) is to find acceleration, and we know from Newton's second law \( F = ma \), so \( a=\frac{F_{net}}{m} \). Wait, first let's get the values: \( F_{EM}=1.98\times 10^{20}\ N \), \( F_{SM}=4.36\times 10^{20}\ N \), mass of moon \( m = 7.35\times 10^{22}\ kg \).

First, calculate the net force. Since the forces are perpendicular, \( F_{net}=\sqrt{F_{EM}^2+F_{SM}^2} \)? Wait no, wait the diagram shows \( F_{EM} \) and \( F_{SM} \) as perpendicular? Wait the problem says "the gravitational force on the Moon due to the Earth, \( \vec{F}_{EM} \), is perpendicular to the gravitational force on the Moon due to the Sun, \( \vec{F}_{SM} \)". So they are perpendicular, so net force magnitude is \( \sqrt{F_{EM}^2 + F_{SM}^2} \)? Wait but in part (a) the user had a wrong expression \( a=\sqrt{(2F_{EM} + 2F_{SM})^{0.5}} \) which is incorrect. Let's do it correctly.

First, compute \( F_{net}=\sqrt{F_{EM}^2 + F_{SM}^2} \)

\( F_{EM}=1.98\times 10^{20}\ N \), \( F_{SM}=4.36\times 10^{20}\ N \)

So \( F_{net}=\sqrt{(1.98\times 10^{20})^2+(4.36\times 10^{20})^2} \)

Factor out \( 10^{20} \):

\( F_{net}=10^{20}\sqrt{(1.98)^2+(4.36)^2} \)

Calculate \( (1.98)^2 = 3.9204 \), \( (4.36)^2 = 19.0096 \)

Sum: \( 3.9204 + 19.0096 = 22.93 \)

So \( F_{net}=10^{20}\sqrt{22.93} \approx 10^{20}\times 4.7885 \approx 4.7885\times 10^{20}\ N \)

Now, Newton's second law: \( a=\frac{F_{net}}{m} \)

\( m = 7.35\times 10^{22}\ kg \)

So \( a=\frac{4.7885\times 10^{20}}{7.35\times 10^{22}} \)

Simplify the exponents: \( 10^{20 - 22}=10^{-2} \)

So \( a=\frac{4.7885}{7.35}\times 10^{-2} \approx 0.6515\times 10^{-2} = 6.515\times 10^{-3}\ m/s^2 \)

Wait, let's check the calculation again. Wait maybe I made a mistake in the net force. Wait the problem's part (a) had a wrong expression, but let's re-express. Wait, maybe the forces are not perpendicular in the way I thought? Wait the diagram shows \( F_{EM} \) and \( F_{SM} \) as two perpendicular vectors, so the net force is the hypotenuse. But let's recalculate \( (1.98\times 10^{20})^2 = (1.98)^2\times 10^{40}=3.9204\times 10^{40} \), \( (4.36\times 10^{20})^2=(4.36)^2\times 10^{40}=19.0096\times 10^{40} \). Sum is \( (3.9204 + 19.0096)\times 10^{40}=22.93\times 10^{40} \). Square root is \( \sqrt{22.93}\times 10^{20}\approx 4.7885\times 10^{20}\ N \). Then divide by mass \( 7.35\times 10^{22}\ kg \): \( \frac{4.7885\times 10^{20}}{7.35\times 10^{22}}=\frac{4.7885}{7.35}\times 10^{-2}\approx 0.6515\times 10^{-2}=6.515\times 10^{-3}\ m/s^2 \approx 6.52\times 10^{-3}\ m/s^2 \) or approximately \( 0.0065\ m/s^2 \).

Wait, maybe I messed up the net force formula. Wait, the problem's part (a) had a wrong expression, but let's check the problem statement again. Wait the user's part (a) says "Write an expression for the magnitude of the acceleration of the Moon. \( a=\sqrt{(2F_{EM} + 2F_{SM})^{0.5}} \) X Incorrect!". So that expression is wrong. The correct approach is: since the two forces are perpendicular, the net force is \( F_{net}=\sqrt{F_{EM}^2 + F_{SM}^2} \), then by Newton's second law \( a = \frac{F_{net}}{m} \), so \( a=\frac{\sqrt{F_{EM}^2 + F_{SM}^2}}{m} \).

Let's plug in the numbers:

\( F_{EM}=1.98\times 10^{20}\ N \), \( F_{SM}=4.36\times 10^{20}\ N \), \( m = 7.35\times 10^{22}\ kg \)

First, compute \( F_{EM}^2=(1.98\times 10^{20})^2 = 3.9204\times 10^{40}\ N^2 \)

\( F_{SM}^2=(4.36\times 10^{20})^2 = 19.0096\times 10^{40}\ N^2 \)

Sum: \( 3.9204 + 19.0096 = 22.93 \), so \( F_{net}^2 = 22.93\times 10^{40}\ N^2 \), so \( F_{net}=\sqrt{22.93}\times 10^{20}\ N \approx 4.7885\times 10^{20}\ N \)

Then \( a=\frac{4.7885\times 10^{20}\ N}{7.35\times 10^{22}\ kg} \)

Calculate the numerical part: \( \frac{4.7885}{7.35} \approx 0.6515 \)

Exponent part: \( 10^{20 - 22}=10^{-2} \)

So \( a \approx 0.6515\times 10^{-2}\ m/s^2 = 6.515\times 10^{-3}\ m/s^2 \approx 6.52\times 10^{-3}\ m/s^2 \) or \( 0.0065\ m/s^2 \) (rounded to two decimal places in the thousandth place).

Step2: Compute the acceleration

Using the formula \( a = \frac{\sqrt{F_{EM}^2 + F_{SM}^2}}{m} \), substitute the given values:

\( F_{EM} = 1.98 \times 10^{20}\ N \), \( F_{SM} = 4.36 \times 10^{20}\ N \), \( m = 7.35 \times 10^{22}\ kg \)

First, calculate the sum of squares inside the square root:

\( F_{EM}^2 + F_{SM}^2 = (1.98 \times 10^{20})^2 + (4.36 \times 10^{20})^2 = 10^{40} \times (1.98^2 + 4.36^2) = 10^{40} \times (3.9204 + 19.0096) = 10^{40} \times 22.93 \)

Take the square root:

\( \sqrt{F_{EM}^2 + F_{SM}^2} = \sqrt{22.93} \times 10^{20} \approx 4.7885 \times 10^{20}\ N \)

Now divide by mass:

\( a = \frac{4.7885 \times 10^{20}}{7.35 \times 10^{22}} = \frac{4.7885}{7.35} \times 10^{-2} \approx 0.6515 \times 10^{-2} = 6.515 \times 10^{-3}\ m/s^2 \approx 6.52 \times 10^{-3}\ m/s^2 \) (or \( 0.0065\ m/s^2 \) when rounded appropriately)

Answer:

\( \approx 6.52 \times 10^{-3}\ m/s^2 \) (or \( 0.0065\ m/s^2 \))