# Physic Question (1 Viewer)

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#### Hybr!d

[SOLVED] Physic Question

1. The problem statement, all variables and given/known data

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. 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.

The mass of the Moon is 7.35×10^22 kg, the mass of the Earth is 6.00×10^24 kg, and the mass of the sun is 2.00×10^30 kg. The distance between the Moon and the Earth is 3.80×10^5 km. The distance between the Earth and the Sun is 1.50×10^8 km.

2. Relevant equations
I am using the x(cm) = m1x1 + m2x2 + m3x3/ m1 + m2 + m3

Where m is masses of the objects and x is distance

3. The attempt at a solution

I put in x(cm) = ((2.00 x 10^30)(1.50×10^8 km) + (7.35×10^22 kg)(3.80×10^5 km) + (6.00×10^24 kg)(1.50×10^8 km))/(7.35×10^22 kg + 6.00×10^24 kg + 2.00×10^30 kg)

I know I am subbing in the wrong value somewhere.

Last edited:

x(cm) = m1x1 + m2x2 + m3x3/ m1 + m2 + m3
If you put x1,x2,x3 in Km (as given in the question) then you will get x in Km not cm

#### Doc Al

Mentor
2. Relevant equations
I am using the x(cm) = m1x1 + m2x2 + m3x3/ m1 + m2 + m3
Nothing wrong with this formula.
I am completely stumped. Like i try to sub the values into the equation but I get a huge number and it doesnt seem right
Show exactly what values you plugged in.

#### Doc Al

Mentor
If you put x1,x2,x3 in Km (as given in the question) then you will get x in Km not cm
In this context, "cm" stands for center of mass (not centimeters).

#### Hybr!d

Nothing wrong with this formula.

Show exactly what values you plugged in.
I put in x(cm) = ((2.00 x 10^30)(1.50×10^8 km) + (7.35×10^22 kg)(3.80×10^5 km) + (6.00×10^24 kg)(1.50×10^8 km))/(7.35×10^22 kg + 6.00×10^24 kg + 2.00×10^30 kg)

#### Doc Al

Mentor
I put in x(cm) = ((2.00 x 10^30)(1.50×10^8 km) + (7.35×10^22 kg)(3.80×10^5 km) + (6.00×10^24 kg)(1.50×10^8 km))/(7.35×10^22 kg + 6.00×10^24 kg + 2.00×10^30 kg)
Measure all positions with respect to the sun, which is at x = 0. What's the distance of each body from the sun?

(I failed to notice that you gave these details in your first post. D'oh!)

#### Hybr!d

Thanks for the help I got it :) Its just

x(cm) = ((2.00 x 10^30)(0) + (7.35×10^22 kg)(3.80×10^5 km + 1.5x10^8) + (6.00×10^24 kg)(1.50×10^8 km))/(7.35×10^22 kg + 6.00×10^24 kg + 2.00×10^30 kg)

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