I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = 
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
Kepler's first law - sometimes referred to as the law of ellipses - explains that planets are orbiting the sun in a path described as an ellipse. An ellipse can easily be constructed using a pencil, two tacks, a string, a sheet of paper and a piece of cardboard. Tack the sheet of paper to the cardboard using the two tacks. Then tie the string into a loop and wrap the loop around the two tacks. Take your pencil and pull the string until the pencil and two tacks make a triangle (see diagram at the right). Then begin to trace out a path with the pencil, keeping the string wrapped tightly around the tacks. The resulting shape will be an ellipse. An ellipse is a special curve in which the sum of the distances from every point on the curve to two other points is a constant. The two other points (represented here by the tack locations) are known as the foci of the ellipse. The closer together that these points are, the more closely that the ellipse resembles the shape of a circle. In fact, a circle is the special case of an ellipse in which the two foci are at the same location. Kepler's first law is rather simple - all planets orbit the sun in a path that resembles an ellipse, with the sun being located at one of the foci of that ellipse.
Answer:
Answer for the question is given in the attachment.
Explanation:
False. C + O --> CO not CO2. Carbonmonoxide
Answer:
The ratio is 
Explanation:
From the question we are told that
The radius of Phobos orbit is R_2 = 9380 km
The radius of Deimos orbit is 
Generally from Kepler's third law
Here M is the mass of Mars which is constant
G is the gravitational constant
So we see that 
=> ![[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%5E2%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E3)
Here 
is the period of Deimos
and is the period of Phobos
So
![[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D)
=> ![\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7BT_1%7D%7BT_2%7D%20%20%3D%20%20%5B%5Cfrac%7B23500%20%7D%7B9380%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D)
=>
