Explanation:
It is given that,
A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.
Mass of the Sun, 
Radius of Mercury's orbit, 
Radius of discovered planet, 

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :




T = 4135214.625 s
or
T = 47.86 days
So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.
Answer:
- The procedure is: solve the quadratic equation for
.
Explanation:
This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:
That is a quadratic equation, where the independent variable is the time
.
Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for
.
To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:
Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:
Where:

That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.
Answer:

Explanation:
The differential electric field
due to differential charge
at distance
from the origin is

but since
we have


integrating this from
to
we get

![E = 4k[(x_0+L)-x_0]](https://tex.z-dn.net/?f=E%20%3D%204k%5B%28x_0%2BL%29-x_0%5D)

putting in
and
we get
