# Calculating forces between stars?

• Tim Wellens
In summary, the forces on the small star are F1 (6.66 x 10^25 N) and F2 (1.33 x 10^26 N). The x and y components of the total force on the small star are x= 2.00 x 10^12 m and y= 0.
Tim Wellens

## Homework Statement

Picture- https://www.flickr.com/photos/137149410@N02/shares/32x7oM

The stars are at vertices of a 45 degree right angle. It is assumed that the stars are spherical so that we can replace each star by a point mass at it's center, as seen in the picture.

1) The two forces acting on the small star are F1 (force due to the upper large star) and F2 (force due to the lower large star). Calculate the magnitude of each force

2) Find the x and y components of the two forces

3) Find the x and y component of the total force on the small star.

F=Gm1m2/r^2[/B]

## The Attempt at a Solution

a) For F1 I used F=Gm1m2/r^2. I plugged in the mass of the small star as mass 1 and the mass of the large upper star as mass 2. I used the distance from the small star to the large star as the r value, which i calculated with pythagorean theorem from the two sides of the right triangle. I got a value of 2.83 x 10^12 m, which I think is correct. So, for F1 I got 6.66 x 10^25 N

For F2, I did the same, except for the r value I used the distance between the small star and the lower large star, which is 2.00 x 10^12 m. The answer i got for F2 was 1.33 x 10^26 N

b) F1x= 2.00 X10^12m F1y= 2.00 x 10^12m
F2x=2.00 x 10^12 m F2y=0

c)I'm not sure on how to calculate the total force on the small star, which I believe is the F vector pointing through the right triangle. I know that Fx= 2.00 x 10^12 m but I'm not sure how to get the Fy value with only a right angle and an x value??[/B]

You did fine for part (a).

In part (b) they're looking for the components of the forces but you supplied the components of the distances. Forces have the units N (Newtons).

For part (c) you'll use the components found in part (b) to add up the force vectors.

So to find the components of the forces.. For F1... Would we use the triangle formed by F1, F, and the dashed line above? So F1y= 6.66 x 10^26 N and F1x= 1.33 x 10^26 N because F1x would be equivalent to F2? Or am i really mucking this up?

Tim Wellens said:
So to find the components of the forces.. For F1... Would we use the triangle formed by F1, F, and the dashed line above? So F1y= 6.66 x 10^26 N and F1x= 1.33 x 10^26 N because F1x would be equivalent to F2? Or am i really mucking this up?

F1 and F2 are entirely independent forces. You want to find the x and y components of both of them.

Note that F1 lies along the line between the small star and the upper large star. So it must have the same angle with respect to the horizontal as that line. You can use that angle and appropriate trig functions to find its components. Or use similar triangles, since you know the lengths of all the sides of the distance triangle, and the F1 lies along the hypotenuse of that triangle:

Note how triangle Oab is similar to triangle OAB.

Ok, so for F1x i would be cos45= f1x/6.66x10^25 which would equal 4.71 x 10^27. F1y was the same due to the angle being 45 degrees. Is this right?

If so, am i then supposed to use the law of cosines to find what the total force is?

Tim Wellens said:
Ok, so for F1x i would be cos45= f1x/6.66x10^25 which would equal 4.71 x 10^27. F1y was the same due to the angle being 45 degrees. Is this right?
No, a component should not be larger than the vector's magnitude. Must be a calculation slip, because your expression looks okay. Try that again. You're right that the two components of F1 should have the same size thanks to the 45° angle.
If so, am i then supposed to use the law of cosines to find what the total force is?
No need. Find the x and y components of both F1 and F2, then add the vectors together. The total force is the sum of the forces.

The reason for breaking the two forces down into components is that it makes adding the vectors together simple.

Yes. I meant to say that I got 4.71 x 10^25 for f1x and f1y. For f2x I'd have 1.33 x 10^26 because there is only an x component of this force.

So for the total force my Fx would be 4.71 x 10^25 (granted that was correct) + 1.33x 10^26 = 1.80 x 10^26 N
Fy= 4.71 x 10^25 N

And the magnitude of this would be gotten via the pythagorean theorem i think.. So F= √(e.80 x 10^26)^2 + (4.71 x 10^25) ^2= 1.86 x 10^26 N

Yup. Looks good.

You might want to keep extra digits in intermediate results. Don't round or truncate values that you'll be using for further calculations, otherwise rounding and truncation errors will creep into significant figures as you proceed. Only round final values to the correct number of significant digits at the end of the calculations.

## 1. How do you calculate the force of gravity between two stars?

The force of gravity between two stars can be calculated using Newton's Law of Universal Gravitation: F = (G * m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two stars, and r is the distance between them.

## 2. What is the significance of calculating forces between stars?

Calculating forces between stars helps us understand the dynamics of celestial bodies, such as their orbits and interactions. It also allows us to make predictions about the future movements and behaviors of stars.

## 3. Can forces between stars change over time?

Yes, forces between stars can change over time. This is due to various factors such as changes in mass, distance, and gravitational pull from surrounding objects.

## 4. How do you account for the effects of dark matter when calculating forces between stars?

Dark matter is a theoretical form of matter that does not interact with light and is difficult to detect. When calculating forces between stars, scientists account for the effects of dark matter by including it as a factor in their calculations. However, the exact nature and behavior of dark matter are still not fully understood.

## 5. Are there different methods for calculating forces between stars?

Yes, there are different methods for calculating forces between stars, depending on the specific scenario and the level of accuracy required. Some methods include using numerical simulations, approximations, or advanced mathematical models.

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