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

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

\(\displaystyle F=ma\)

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

\(\displaystyle G\frac{m*m_{E}}{r^2}=mg\)

Notice that the mass cancels out from both sides.

\(\displaystyle G\frac{m_{E}}{r^2}=g\)

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

The new planet has a mass equal to twice that of Earth. That means it has a mass of \(\displaystyle 2m_E\). It also has half the radius of Earth, \(\displaystyle \small \frac{1}{2}r\). Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

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

Expand this equation in order to combine the non-variable terms.

\(\displaystyle G\frac{2m_{E}}{\frac{1}{4}r^2}=a\)

\(\displaystyle \frac{2}{\frac{1}{4}}(G\frac{m_{E}}{r^2})=a\)

\(\displaystyle 8(G\frac{m_{E}}{r^2})=a\)

We had previously solved for the gravity on Earth:

\(\displaystyle G\frac{m_{E}}{r^2}=g\)

We can substitute this into the new acceleration equation:

 \(\displaystyle 8g=a\)

The acceleration due to gravity on this new planet will be eight times what it would be on Earth.

For this comparison we can use the law of universal gravitation:

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

The astronaut notices that the force due to gravity on this planet is half what it is on Earth. We can thus relate the force equations for the two planets.

\(\displaystyle F_{p}=\frac{1}{2}F_G\)

\(\displaystyle \frac{1}{2}(G\frac{m_1m_E}{r^2})=G\frac{m_1m_p}{r^2}\)

\(\displaystyle m_1\) refers to any arbitrary mass on the surface of the planet, and will be constant. The radii of the two planets are equal, and the gravitational constant is also equal. These three variables can cancel out from both sides of the equation.

\(\displaystyle \frac{1}{2}m_E(G\frac{m_1}{r^2})=m_p(G\frac{m_1}{r^2})\)

\(\displaystyle \frac{1}{2}m_E=m_p\)

This relation shows us that the mass of the new planet will be half that of Earth.

For this comparison we can use the law of universal gravitation:

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

The astronaut notices that the force due to gravity on this planet is twice what it is on Earth. We can thus relate the force equations for the two planets.

\(\displaystyle F_{p}=2F_G\)

\(\displaystyle 2(G\frac{m_1m_E}{r^2})=G\frac{m_1m_p}{r^2}\)

\(\displaystyle m_1\) refers to any arbitrary mass on the surface of the planet, and will be constant. The radii of the two planets are equal, and the gravitational constant is also equal. These three variables can cancel out from both sides of the equation.

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

\(\displaystyle 2m_E=m_p\)

This relation shows us that the mass of the new planet will be twice that of Earth.

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.

\(\displaystyle m_1=1kg;\ m_2=1kg;\ r=1m\)

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

\(\displaystyle \small G\) 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.

\(\displaystyle m_{1a}=2m_1=2kg;\ m_{2a}=3m_2=3kg;\ r_a=4r=4m\)

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

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

\(\displaystyle F_2=\frac{3}{8}G\)

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

\(\displaystyle F_1=G\ \text{and}\ F_2=\frac{3}{8}G\)

\(\displaystyle F_2=\frac{3}{8}F_1\)

To solve this problem, 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.

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

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

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

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

\(\displaystyle F_G=6.87N\)

This force will be the same for both objects. The Earth will exert the same force on the satellite as the satellite exerts on the Earth. This makes sense, given Newton's third law.

To solve this problem, 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.

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

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

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

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

\(\displaystyle F_G=6.87N\)

This force will be the same for both objects. The Earth will exert the same force on the satellite as the satellite exerts on the Earth. This makes sense, given Newton's third law.

To solve this problem, first 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!

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

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

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

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

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

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

To solve this problem, 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 mass of the Earth, force of gravity, and distance to solve for the mass of the satellite.

\(\displaystyle 7.99N=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(m_s)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}\)

\(\displaystyle 7.99N=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(m_s)(5.97*10^{24}kg)}{1.45*10^{17}m^2}\)

\(\displaystyle \frac{7.99N}{6.67*10^{-11}\frac{m^3}{kg*s^2}}=\frac{(m_s)(5.97*10^{24}kg)}{1.45*10^{17}m^2}\)

\(\displaystyle 1.198*10^{11} \frac{kg^2}{m^2}=\frac{(m_s)(5.97*10^{24}kg)}{1.45*10^{17}m^2}\)

\(\displaystyle (1.198*10^{11} \frac{kg^2}{m^2})*(1.45*10^{17}m^2)={(m_s)(5.97*10^{24}kg)}\)

\(\displaystyle 1.74*10^{28}kg^2={(m_s)(5.97*10^{24}kg)}\)

\(\displaystyle \frac{1.74*10^{28}kg^2}{5.97*10^{24}kg}={m_s}\)

\(\displaystyle 2914.6kg=m_s\)

To solve this problem, we need to set up the law of universal gravitation.

When the satellite was whole:

\(\displaystyle F_{G}_1=G\frac{m_Em_s}{r^2}\)

After the satellite breaks:

\(\displaystyle F_G_2=G\frac{m_E\frac{1}{2}m_s}{r^2}\)

We can now compare these two equations. Start by expanding the second equation.

\(\displaystyle F_G_2=G\frac{m_E\frac{1}{2}m_s}{r^2}=\frac{1}{2}*G\frac{m_Em_s}{r^2}\)

We can substitute our first equation into the second.

\(\displaystyle F_{G}_1=G\frac{m_Em_s}{r^2}\) and \(\displaystyle F_G_2=\frac{1}{2}*G\frac{m_Em_s}{r^2}\)

\(\displaystyle F_G_2=\frac{1}{2}*F_G_1\)

This tells us that the final force of gravity is equal to one half the original force of gravity.

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.

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

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

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

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

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