In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate the center of mass of the Earth-Moon-Sun system. The mass of the Moon is 7.35×1022 kg , the mass of the Earth is 6.00×1024 kg , and the mass of the sun is 2.00×1030 kg . The distance between the Moon and the Earth is 3.80×105 km . The distance between the Earth and the Sun is 1.50×108 km

A.) Where is the center of mass of the Earth-Moon system? The radius of the Earth is 6378 km and the radius of the Moon is 1737 km. Select one of the answers below:

a. The center of mass is exactly in the center between the Earth and the Moon.
b. The center of mass is nearer to the Moon than the Earth, but outside the radius of the Moon.
c. The center of mass is nearer to the Earth than the Moon, but outside the radius of the Earth.
d. The center of mass is inside the Earth.
e. The center of mass is inside the Moon.
f.) Calculate the location of the center of mass of the Earth-Moon-Sun system during a full Moon. A full Moon occurs when the Earth, Moon, and Sun are lined up as shown in the figure. (Figure 2) Use a coordinate system in which the center of the sun is at x=0 and the Earth and Moon both lie along the positive x direction.

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

a) 37 N

b) 2.5 m/s²

Explanation:

Let's say m₁ is the 5 kg mass, and m₂ is the 3 kg mass.

Draw a free body diagram for each block.  For both, there are two forces: weight down and tension up.

Apply Newton's second law to the 5 kg mass in the y direction (remember it is accelerating downward).

∑F = ma

T − m₁g = m₁(-a)

m₁g − T = m₁a

Apply Newton's second law to the 3 kg mass in the y direction:

∑F = ma

T − m₂g = m₂a

a) Solve the system of equations.  First, solve for a in each equation, then set equal and solve for T.

a = (m₁g − T) / m₁

a = (T − m₂g) / m₂

(m₁g − T) / m₁ = (T − m₂g) / m₂

m₂ (m₁g − T) = m₁ (T − m₂g)

m₁m₂g − Tm₂ = m₁T − m₁m₂g

2m₁m₂g = (m₁ + m₂)T

T = 2m₁m₂g / (m₁ + m₂)

Given m₁ = 5 kg and m₂ = 3 kg:

T = 2 (5 kg) (3 kg) (9.8 m/s²) / (5 kg + 3 kg)

T = 36.75 N

Rounded to two significant figures, the tension is 37 N.

b) Solve the system of equations.  First, solve for T in each equation, then set equal and solve for a.

T = m₁g − m₁a

T = m₂g + m₂a

m₁g − m₁a = m₂g + m₂a

m₁g − m₂g = m₁a + m₂a

(m₁ − m₂)g = (m₁ + m₂)a

a = (m₁ − m₂)g / (m₁ + m₂)

Given m₁ = 5 kg and m₂ = 3 kg:

a = (5 kg − 3 kg) (9.8 m/s²) / (5 kg + 3 kg)

a = 2.45 m/s²

Rounded to two significant figures, the magnitude of the acceleration of the blocks is 2.5 m/s².