Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orbit, assuming

Question

Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orbit, assuming a circular orbit?

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Amity 3 months 2021-08-15T08:07:14+00:00 1 Answers 11 views 0

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    2021-08-15T08:08:46+00:00

    Answer: 58,045,522,878.8 meters

    Explanation:

    Ok, the data we have is

    Period = T = 88 days

    Radial acceleration = ar = 3.96×10^-2 m/s^2

    And we know that the equation for the radial acceleration is:

    ar = v^2/r = r*w^2

    Where v is the velocity. r is the radius and w is the angular velocity.

    And we know that:

    w = 2*pi*f

    where f is the frequency, and:

    T = 1/f.

    Then we can write:

    w = 2*pi/T

    and our equation becomes:

    ar = r*(2*pi/T)^2

    Now we solve this for r.

    First we need to use the same units in both equations, so we want to write T in seconds.

    T = 88 days,

    A day has 24 hours, and one hour has 3600 seconds:

    T = 88*24*3600 s =7,603,200s

    Then:

    3.96×10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

    r = (3.96×10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

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