Kinetic friction is never greater than static friction. More force is always requires to overcome static friction than is required to overcome kinetic friction. It can require a large force to initiate motion, causing an initial acceleration by overcoming static friction. Once motion has begun, however, less for is required to maintain the motion due to the principles of Newton's first law and inertia.

The relationship between force and acceleration is Newton's second law:

\(\displaystyle F=ma\)

We know the mass, but we will need to find the force. For this calculation, use the law of universal gravitation:

\(\displaystyle F=\frac{Gm_1m_2}{r^2}\)

We are given the value of each mass, the distance (radius), and the gravitational constant. Using these values, we can solve for the force of gravity.

\(\displaystyle F=\frac{(6.67x10^{-11}m^3/kgs^2)(10x10^{15}kg)(5.972x10^{24}kg)}{(250,000,000m)^2}\)

\(\displaystyle F=\frac{(6.67x10^{-11}m^3/kgs^2)(5.972)(1040kg)}{26.25x10^{16}m^2}\)

\(\displaystyle F=\frac{(6.67x10^{-11}m^3/kgs^2)(9.5552x10^{23}kg^2m^2)}{?}\)

\(\displaystyle F=6.37x10^{13}N\)

Now that we know the force, we can use this value with the mass of the asteroid to find its acceleration.

\(\displaystyle F=ma\)

\(\displaystyle 6.37x10^{13}N=(10x10^{15}kg)a\)

\(\displaystyle \frac{6.37x10^{13}N}{10x10^{15}kg}=a\)

\(\displaystyle 0.00637m/s^2=a\)

The equation for the force of gravity between two objects is:

\(\displaystyle F=\frac{Gm_1m_2}{r^2}\)

Using this equation, we can select arbitrary values for our original masses and distance. This will make it easier to solve when these values change.

\(\displaystyle m1=1kg; m2=1kg; r=1m\)

\(\displaystyle F_1=\frac{G(1kg)(1kg)}{(1m)2}=G\)

\(\displaystyle G\) is the gravitational constant. Now that we have a term for the initial force of gravity, we can use the changes from the question to find how the force changes.

\(\displaystyle m1a=2m1=2kg; m2a=3m2=3kg; ra=4r=4m\)

\(\displaystyle F_2=\frac{G(2kg)(3kg)}{(4m)^2}\)

\(\displaystyle F_2=\frac{6kg^2G}{16m^2}\)

\(\displaystyle F_2=38G\)

We can use our first calculation to see the how the force has changed.

\(\displaystyle F_1=G \: and\: F_2=38G\)

\(\displaystyle F_2=38F_1\)

We can use the conservation of momentum to solve. Since the satellites stick together, there is only one final velocity term.

\(\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_3\)

We know the masses for both satellites are equal, and the second satellite is initially stationary.

\(\displaystyle (2500kg*v_1)+(2500kg*0ms)=(2500kg+2500kg)v_3\)

Now we need to find the velocity of the first satellite. Since the satellite is in orbit (circular motion), we need to find the tangential velocity. We can do this by finding the centripetal acceleration from the centripetal force.

Recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal force acting on the satellite. That means \(\displaystyle Fc=FG\).

Solve for \(\displaystyle FG\) for the satellite. To do this, use the law of universal gravitation.

\(\displaystyle F=\frac{Gm_1m_2}{r^2}\)

Remember that r is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

\(\displaystyle F_G=\frac{Gm_sm_E}{(r_E+h)^2}\)

\(\displaystyle F_G=\frac{(6.67x10^{-11}m^3/kgs)(2500kg)(5.97x10^{24}kg)}{(6.37x10^5m+3.8x10^8m)^2}\)

\(\displaystyle F_G=\frac{(6.67x10^{-11}m^3/kgs^2)(1.49)(1028kg)}{21.45x10^{17}m^2}\)

\(\displaystyle F_G={(6.67x10^{-11}m^3/kgs^2)(1.03x10^{11}kg^2m^2)\)

\(\displaystyle F_G=6.87N\)

Now that we know the force, we can find the acceleration. Remember that centripetal force is Fc=m∗ac. Set our two forces equal and solve for the centripetal acceleration.

\(\displaystyle F_G=F_c=m_sa_c\)

\(\displaystyle 6.87N=m_sa_c\)

\(\displaystyle 6.87N=2500kga_c\)

\(\displaystyle 2.7x10^{-3}m/s^2=a_c\)

Now we can find the tangential velocity, using the equation for centripetal acceleration. Again, remember that the radius is equal to the sum of the radius of the Earth and the height of the satellite!

\(\displaystyle a_c=\frac{v^2}{r}\)

\(\displaystyle 2.7x10^{-3}m/s^2=\frac{v^2}{1.45x10^{17}m}\)

\(\displaystyle (2.7x10^{-3}m/s^2)(1.45x10^{17}m)=v^2\)

\(\displaystyle 2.85x10^{15}m^2/s^2=v^2\)

\(\displaystyle \sqrt{2.85x10^{15}m^2/s^2}\)

\(\displaystyle 5.33x10^7m/s=v\)

This value is the tangential velocity, or the initial velocity of the first satellite. We can plug this into the equation for conversation of momentum to solve for the final velocity of the two satellites.

\(\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_3\)

\(\displaystyle (2500kg)(5.33x10^7m/s)+(2500kg)(0m/s)=(2500kg+2500kg)v_3\)

\(\displaystyle 1.3325x10^{11}kgm/s=(5000kg)v_3\)

\(\displaystyle \frac{1.3325x10^{11}kgm/s}{5000kg}=v_3\)

\(\displaystyle 2.67x10^7m/s=v_3\)

For this comparison, we can use the law of universal gravitation and Newton's second law:

\(\displaystyle F=\frac{Gm_1m_2}{r^2}\)

\(\displaystyle F=ma\)

We know that the force due to gravity on Earth is equal to mg. We can use this to set the two force equations equal to one another.

\(\displaystyle \frac{Gmm_E}{r^2}=mg\)

Notice that the mass cancels out from both sides.

\(\displaystyle \frac{Gm_E}{r^2}=g\)

This equation sets up the value of acceleration due to gravity on Earth.

The new planet has a radius equal to twice that of Earth. That means it has a radius of 2r. It has the same mass as Earth, mE. Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

\(\displaystyle \frac{Gm_E}{(2r)^2}=a\)

Expand this equation to compare it to the acceleration of gravity on Earth.

\(\displaystyle \frac{Gm_E}{4r^2}=a\)

\(\displaystyle \frac{1}{4}\frac{Gm_E}{r^2}=a\)

We had previously solved for the gravity on Earth:

\(\displaystyle \frac{Gm_E}{r^2}=g\)

We can substitute this into the new acceleration equation:

\(\displaystyle \frac{1}{4}g=a\)

The acceleration due to gravity on this new planet will be one quarter of what it would be on Earth.

The formula for force and acceleration is Newton's 2nd law: \(\displaystyle F=ma\). We know the mass, but first we need to find the force:

For this equation, use the law of universal gravitation:

\(\displaystyle F=\frac{Gm_1m_2}{r^2}\)

We know from the first equation that a force is a mass times an acceleration. That means we can rearrange the equation for universal gravitation to look a bit more like that first equation:

\(\displaystyle F=\frac{Gm_1m_2}{r^2}\) can turn into: \(\displaystyle F=\frac{Gm_2m_1}{r^2}\) respectively.

We know that the forces will be equal, so set these two equations equal to each other:

\(\displaystyle F=\frac{Gm_1m_2}{r^2}=\frac{Gm_2m_1}{r^2}\)

The problem tells us that \(\displaystyle m_2=2m_1\)

\(\displaystyle \frac{Gm_1(2m_1)}{r^2}=\frac{G(2m_1)m_1}{r^2}\)

Let's say that \(\displaystyle \frac{Gm_1}{r^2}=a\)  to simplify. 

\(\displaystyle m1(2a)=(2m_1)a\)

As you can see, the acceleration for \(\displaystyle m_1\) is twice the acceleration for \(\displaystyle m_2\). Therefore the mass 2m will have the smaller acceleration.

The space station and the astronauts inside are in a constant state of free fall toward the center of the Earth.  However, because they have such a high horizontal velocity and because the Earth is curved they will always be falling toward the earth as the Earth curves away from them.  IF the space station were to slow down, they would land on the Earth.  The high speed in the horizontal direction, keeps them in a parabolic flight path that aligns with the curvature of the Earth.

To find the relationship described in the question, we need to use the law of universal gravitation:

\(\displaystyle F=\frac{Gm_1m_2}{r^2}\)

The question suggests that we know the radius and one of the masses, and asks us to solve for the other mass.

\(\displaystyle F=\frac{Gm_{astro}m_{planet}}{r^2}\)

Since G is a constant, if we know the mass of the astronaut and the radius of the planet, all we need is the force due to gravity to solve for the mass of the planet. According to Newton's third law, the force of the planet on the astronaut will be equal and opposite to the force of the astronaut on the planet; thus, knowing her force on the planet will allows us to solve the equation.

To solve this problem, use Newton's law of universal gravitation:

\(\displaystyle F=\frac{Gm_1m_2}{r^2}\)

We are given the constant, as well as the satellite masses and distance (radius). Using these values we can solve for the force.

\(\displaystyle F_G=\frac{(6.67x10^{-11}m^3/kgs^2)(1723kg)(1723kg)}{(890m)^2}\)

\(\displaystyle F_G=\frac{(6.67x10^{-11}m^3/kgs^2)(2968729kg^2)}{792100m^2)}\)

\(\displaystyle FG=2.5x10^{-10}N\)

Newton’s 3rd law states that for every force there is an equal and opposite force.  In other words, the force with which the moon pulls on the Earth is the same force that the Earth pulls on the moon.

Newton’s 2nd law states that the acceleration of an object is directly related to the force applied and inversely related to the mass of the object.  Since both the earth and the moon have the same force acting on it, it is their masses that will determine who will accelerate more.  Since there is an inverse relationship between the mass and acceleration, the object with the smaller mass will accelerate more.  Therefore the moon will accelerate more.