Anonymous asked in Science & MathematicsPhysics · 1 decade ago

The Hubble Space Telescope orbits Earth 613 km above Earth's surface.?

What is the Period of the telescopes orbit?

3 Answers

  • 1 decade ago
    Best Answer

    Let telescope hav e a mass m and velocity v.

    Then "centrifugal force" (or whatever you call it) = m v^2/R

    where R = r + 613,000 (r is earth's radius, meters)

    This must equal gravitational attraction. We know that at the Earth's surface this would be equal to mg. We also know that it obeys an inverse square law. So at 613 km, it will be equal to mg x r^2/R^2

    So mv^2/R = mg x r^2/R^2

    m of course cancels out. You need to substitute values of r and g, and do the arithmetic.

    Knowing v, period = 2 pi R/v (length of orbit divided by speed)

    Make sure you understand each point in the reasoning. the secret is that gravitational force has to be equal to centrifugal "force". the rest is just thinking about the geometry, and tedious but straightforward arithmetic.

    Good luck

  • 3 years ago

    Any merchandise in low earth orbit could have an orbital era of roughly ninety minutes. For a much better suited fee, you will could desire to prepare the consumer-friendly gravitational consistent, the mass of the earth, and the radius of the orbit as taken from the earth's center. having reported that, use the gravitational acceleration on earth's floor, calculate the relief in that with the aid of making use of orbit's distance above the outdoors, and use that stress to stability the centrifugal stress springing up from the satellite tv for pc's inertia. The latter computation could be much less complicated with the aid of making use of fact it avoids coping with extensive numbers.

  • 1 decade ago

    1.612 hours

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