
Explanation:
The acceleration due to gravity g is defined as

and solving for R, we find that

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration
experienced by the satellite is equal to the gravitational force
or

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

Solving for <em>M</em>, we get

Putting this expression back into Eqn(1), we get





Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on
Limitations: yield only planet's mass and orbital properties
Answer:
f1= -350cm or -3.5m
f2= 22.1cm or 0.221m
Explanation:
A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.
Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.