Relevant equations:

$\displaystyle E_{total}=K+U_g$

$\displaystyle K = \frac{1}{2}mv^2$

$\displaystyle U_g=\frac{-GMm}{r}$

For the rocket to escape the moon's gravity, its minimum total energy is zero. If the total energy is zero, the rocket will have zero final velocity when it is infinitely far from the moon. If total energy is less than zero, the rocket will fall back to the moon's surface. If total energy is greater than zero, the rocket will have some final velocity when it is infinitely far away. 

For the minimum energy case as the rocket leaves the surface:

$\displaystyle 0 = K + U_g = \frac{1}{2}mv_{esc}^2 - \frac{GMm}{r}$

Rearrange energy equation to isolate the velocity term.

$\displaystyle v_{esc}^2 = \frac{GMm}{\frac{1}{2}mr}$

$\displaystyle v_{esc}=\sqrt{\frac{2GM}{r}}$

Substitute in the given values to solve for the velocity.

$\displaystyle v_{esc}=\sqrt{\frac{2(6.67*10^{-11}\frac{m^2N}{kg^2})(7.2*10^{22}kg)}{(1.8*10^6m)}}=2310\frac{m}{s}\approx 2.3\frac{km}{s}$

Relevant equations:

$\displaystyle |F_g| = \frac{GMm}{r^2}$

Use the given values to solve for the force.

$\displaystyle |F_g| = \frac{(6.67*10^{-11}\frac{m^2N}{kg^2})(2.0*10^{30}kg)(5kg)}{(1.5*10^{11}m)^2}=0.0296N\approx 3.0*10^{-2}N$

Newton's law of universal gravitation states:

$\displaystyle F_g=\frac{Gm_1 m_2}{r^2}$

We can write two equations for the gravity experienced before and after the doubling:

$\displaystyle F_{g,1} =\frac{Gm_1 m_2}{d^2}$

$\displaystyle F_{g,2}=\frac{Gm_1 m_2}{(2d)^2}$

The equation for gravity after the doubling can be simplified:

$\displaystyle F_{g,2}=\frac{Gm_1 m_2}{4d^2}$

$\displaystyle F_{g,2}=\frac{1}{4}\frac{Gm_1 m_2}{d^2}$

Because the masses of the spheres remain the same, as does the universal gravitation constant, we can substitute the definition of Fg1 into that equation:

$\displaystyle F_{g,2} = \frac{1}{4}F_{g,1}$

The the gravitational force decreases by a factor of 4 when the distance between the two spheres is doubled.

Newton's law of universal gravitation states that:

$\displaystyle F_g=\frac{Gm_1m_2}{r^2}$

We can write two equations representing the force of gravity before and after the doubling of the mass:

$\displaystyle F_{g,1}=\frac{Gm_{a,1}m_{b,1}}{r^2}$

$\displaystyle F_{g,2}=\frac{Gm_{a,2}m_{b,2}}{r^2}$

The problem gives us $\displaystyle m_{a,2}=2m_{a,1}$ and we can assume that all other variables stay constant.

Substituting these defintions into the second equation:

$\displaystyle F_{g,2}=\frac{G(2m_{a,1})m_{b,1}}{r^2}$

This equation simplifies to:

$\displaystyle F_{g,2}=2 \frac{Gm_{a,1}m_{b,1}}{r^2}$

Substituting the definition of Fg1, we see:

$\displaystyle F_{g,2}=2F_{g,1}$

Thus the gravitational forces doubles when the mass of one object doubles.

When the elevator accelerates upward, we know that an object would appear heavier. The normal force is the sum of all the forces added up, and in this case it is $\displaystyle 35\:N$. We know that the normal force has two components, a component from gravity, and a component from the acceleration of the elevator. Using this equation, we can determine the mass of the block, which doesn't change:

$\displaystyle F_n = ma + mg = m(g+a)$

$\displaystyle g$ is acceleration due to gravity and $\displaystyle a$ is the acceleration of the elevator.

When substituting in the values, we get $\displaystyle 35 = m(9.8+1.5)$

Solving for $\displaystyle m$, we get $\displaystyle m = 3.097\:kg$

Since the elevator is descending at constant speed, no additional force is applied, therefore the force that the spring scale reads is only due to gravity, which is calculated by:

$\displaystyle f=mg$ 

$\displaystyle 3.097 \cdot 9.8 = 30.35\:N$

To do this problem we have to realize that the force of gravity acting on the bullet is equal to the centripetal force. The equations for gravitational force and centripetal force are as follows:

$\displaystyle F_g = \frac{GmM}{r^2}, F_c = \frac{mv^2}{r}$

If we set the two equations equal to each other, the small $\displaystyle m$ (mass of the bullet) will cancel out and $\displaystyle r$ will disappear from the right side of the equation.

$\displaystyle \frac{GM}{r}= v^2 \rightarrow v=\sqrt{\frac{GM}{r}}$

$\displaystyle G$ is the universal gravitational constant $\displaystyle 6.67384 \times10^{-11}\: m^3 kg^{-1} s^{-2}$

$\displaystyle M$ is given to be $\displaystyle 5.35\times10^{22} kg$ and $\displaystyle r$ to be $\displaystyle 1740\:km \rightarrow 1,740,000\:m$.

If we plug everything in, we get

$\displaystyle 1430\:m/s$ or $\displaystyle 1.43\:km/s$