Acomet goes around the sun in an elliptical orbit. at its farthest point, 490 million miles from the sun, it is traveling with a speed of 15,470 mi/h. how fast is it traveling at its closest approach to the sun, at a distance of 124 million miles?

V = 61,131.45 mi/h

Explanation:

In this case, we need to use the expression for linear momentum which is:

L = mVd

Where:

m = mass of the object

V = speed of the object

d = distance of the object or radius.

We also need to know that the law of conservation of energy is applied here, so:

L1 = L2

We can say that L1 would be the linear momentum of the comet at it's farthest point and L2 would be the linear momentum of the comet at it's closest point. Then, the expressions for linear momentum for both cases are:

m*V1*d1 = m*V2*d2

And now, all we need to do is calculate V2. So solving for V2:

V2 = m*V1*d1 / m*d2

V2 = V1*d1 / d2

Replacing all the data we have:

V2 = 490,000,000 * 15,470 / 124,000,000

V2 = 61,131.45 mi/h

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laticia draws the diagram below to show earth’s magnetic field.

answer

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