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## 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

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## Answers ( )

Answer:58,045,522,878.8 metersExplanation: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