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.

\(\displaystyle (2500kg*v_1)+(2500kg*0\frac{m}{s})=(2500kg+2500kg)v_3\)

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 \(\displaystyle F_c=F_G\).

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

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

Remember that \(\displaystyle r\) 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.

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

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})\)

\(\displaystyle F_G=6.87N\)

Now that we know the force, we can find the acceleration. Remember that centripetal force is \(\displaystyle F_c=m*a_c\). Set our two forces equal and solve for the centripetal acceleration.

\(\displaystyle F_G=F_c=m_s*a_c\)

\(\displaystyle 6.87N=m_s*a_c\)

\(\displaystyle 6.87N=2500kg*a_c\)

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}=a_c\)

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!

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

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}\)

\(\displaystyle (2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}\)

\(\displaystyle 2.85*10^{15}\frac{m^2}{s^2}=v^2\)

\(\displaystyle \sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}\)

\(\displaystyle 5.33*10^7\frac{m}{s}=v\)

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.

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

\(\displaystyle (2500kg*5.33*10^{7}\frac{m}{s})+(2500kg*0\frac{m}{s})=(2500kg+2500kg)v_3\)

\(\displaystyle 1.3325*10^{11}\frac{kg*m}{s}=(5000kg)v_3\)

\(\displaystyle \frac{1.3325*10^{11}\frac{kg*m}{s}}{5000kg}=v_3\)

\(\displaystyle 2.67*10^7\frac{m}{s}=v_3\)

The period describes how long it takes the satellite to make one full orbit. If you go back to the definition of velocity, \(\displaystyle v=\frac{\Delta x}{\Delta t}\), we can apply that to our new circular orbit, in which the distance is equal to the circumference of the circle and the time is equal to the period: \(\displaystyle v=\frac{2\pi r}{T}\). The circumference divided by the period will give us the average velocity.

The problem gives us the radius, but we need to find the tangential velocity. We can do this by first solving for 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 \(\displaystyle F_c=F_G\).

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

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

Remember that \(\displaystyle r\) 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.

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

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})\)

\(\displaystyle F_G=6.87N\)

Now that we know the force, we can find the acceleration. Remember that centripetal force is \(\displaystyle F_c=m*a_c\). Set our two forces equal and solve for the centripetal acceleration.

\(\displaystyle F_G=F_c=m_s*a_c\)

\(\displaystyle 6.87N=m_s*a_c\)

\(\displaystyle 6.87N=2500kg*a_c\)

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}=a_c\)

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!

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

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}\)

\(\displaystyle (2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}\)

\(\displaystyle 2.85*10^{15}\frac{m^2}{s^2}=v^2\)

\(\displaystyle \sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}\)

\(\displaystyle 5.33*10^7\frac{m}{s}=v\)

We now have a value for the tangential velocity, which we can use in the equation for velocity from the beginning to find the period.

\(\displaystyle v=\frac{\Delta x}{\Delta t}=\frac{2\pi r}{T}\)

\(\displaystyle 5.33*10^7\frac{m}{s}=\frac{2\pi(1.45*10^{17}m)}{T}\)

\(\displaystyle T=\frac{2\pi(1.45*10^{17}m)}{5.33*10^7\frac{m}{s}}\)

\(\displaystyle T=1.71*10^{11}s\)

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

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

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

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}\)

\(\displaystyle F_G=1.24*10^{20}N\)

Remember that Newton's second law states that \(\displaystyle F=ma\). The force acting upon the planet in question will be the force due to gravity. Once we find that, we can find the acceleration.

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

\(\displaystyle F=G\frac{m_1*m_2}{r^2}\)

Remember that the weight of the astronaut is the same as the gravitational force acting between the planet and the astronaut.

\(\displaystyle F_g=mg=G\frac{m_1*m_2}{r^2}\)

We are given the gravitational constant, the radius of the planet, the mass of the astronaut, and the magnitude of the force generated. Using these values in the universal gravitation equation, we can solve for the mass of the planet.

\(\displaystyle 510N=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(71kg)m_2}{(6*10^7m)^2}\)

\(\displaystyle \frac{510N}{6.67*10^{-11}\frac{m^3}{kg*s^2}}=\frac{(71kg)m_2}{(6*10^7m)^2}\)

\(\displaystyle 7.65*10^{12}\frac{kg^2}{m^2}=\frac{(71kg)m_2}{3.6*10^{15}m^2}\)

\(\displaystyle (7.65*10^{12}\frac{kg^2}{m^2})(3.6*10^{15}m^2)=(71kg)m_2\)

\(\displaystyle 2.75*10^{28}kg^2=(71kg)m_2\)

\(\displaystyle 3.87*10^{26}=m_2\)

We are given the weight of a person on Earth in units of Newtons, which means we can recognize this as a force. The force that is acting on this person is the force due to gravity, which can be represented by the following equation:

\(\displaystyle F=G\frac{m_{1}m_{2}}{r^{2}}\)

\(\displaystyle G\) is the universal gravitational constant and is equal to \(\displaystyle 6.67384*10^{-11} \frac{m^{3}}{kg\cdot s^{2}}\)

\(\displaystyle m_{1}\) is the mass of object 1

\(\displaystyle m_{2}\) is the mass of object 2

\(\displaystyle r\) is the distance between the centers of the two objects

It's important to note that an object's mass will stay the same no matter where it is, but its weight will vary depending on where it is measured. Notice that when calculating the gravitational force, we need to consider the mass of two objects. If we set the mass of the Earth and the mass of the person in question as the two masses, we can rewrite the equation as:

\(\displaystyle F_{Earth}=G\frac{m_{person}m_{Earth}}{r^{2}}=800N\)

To calculate how much the person weighs on the new planet, we need to consider the information given - that the new planet is twice as dense as Earth. This means that for a given volume, the new planet will have twice as much mass as Earth. Furthermore, we know that the mass of the person stays the same since, as mentioned above, mass is constant no matter where it is measured. And, if we are considering a case where the volume is the same, then the distance between the centers should also be the same. Thus, we can calculate the new force as:

\(\displaystyle F_{new\: planet}=G\frac{m_{person}(2m_{Earth})}{r^{2}}=2F_{Earth}=2\left ( 800 N\right )=1600\:N\)

This problem can be solved by looking at the formula for the gravitational constant, \(\displaystyle g\).

\(\displaystyle g=G\frac{M_E}{R_E^2}\)

We are looking for the altitude above the Earth where the gravitation field is \(\displaystyle \frac{1}{9}\) in strength, so we proceed as follows

\(\displaystyle \frac{g}{9}=\frac{GM_E}{\left(dR_E \right )^2}\Rightarrow d=3R_E\Rightarrow r=d-R_E=2R_E\)

Use Newton's law of gravitation to make sense of this problem:

\(\displaystyle F=\frac{Gm_{1}m_{2}}{r^2}\)

As you can see the force of gravity is inversely proportional to the square of the radius. Since \(\displaystyle P_{1}\) has a radius 6 times larger than \(\displaystyle P_{2}\), we state that \(\displaystyle g_{1}=\frac{g_{2}}{36}\) because the acceleration of the gravity of planet two divided by 36 would equal the smaller gravitational acceleration of planet one. 

Universal Gravitation law:

\(\displaystyle F_{grav}=-G\frac{m_1m_2}{r^2}\)

Assume \(\displaystyle m_1\) is the Moon, and \(\displaystyle m_2\) is the object on the surface, then:

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

Will be analogous to \(\displaystyle g\) in \(\displaystyle F_{grav}=mg\), which will be a negative number since it is pointing down.

Plug in values:

\(\displaystyle -6.674*10^{-11}\frac{7.34*10^{22}}{(1,740*10^3)^2}=g_{moon}\)

\(\displaystyle g_{moon}=1.62\frac{m}{s^2}\)