Center of Mass of Earth-Moon-Sun System During Full Moon

[Hybr!d]

[SOLVED] Physic Question

Homework Statement

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.

Homework Equations

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

Where m is masses of the objects and x is distance

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.

 

[Google_Spider]

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]

Homework 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 doesn't seem right

Show exactly what values you plugged in.

 

[Doc Al]

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]

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)