The strength of the gravitational field is given by:

where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:

And now we can use the previous equation to calculate the field strength at that altitude:

And we can see this value is a bit less than the gravitational strength at the surface, which is

.
Answer:
A. increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object
Explanation:
Because the tip of the moon's shadow ... the area of "totality" ... is never more than a couple hundred miles across, It never covers a single place for more than 7 minutes, and can never stay on the Earth's surface for more than a few hours altogether during one eclipse.
If you're not inside that small area, you don't see a total eclipse.
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:

where q is the charge of the proton,

, with

being the elementary charge, and

and

are the initial and final voltage.
Substituting, we get (in electronvolts):

and in Joule: