Given that Newton's second law is \(\displaystyle F=ma\), we can find the acceleration by first determining the force.

To find the gravitational force, use Newton's law of universal gravitation:

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

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

\(\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})\)

\(\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})\)

\(\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})\)

\(\displaystyle F_G=2.53*10^{-5}N\)

We now have values for both the mass and the force. Using the original equation, we can now solve for the acceleration.

\(\displaystyle F=ma\)

\(\displaystyle 2.53*10^{-5}N=(5.33*10^8kg)a\)

\(\displaystyle \frac{2.53*10^{-5}N}{5.33*10^8kg}{}=a\)

\(\displaystyle 4.74*10^{-14}\frac{m}{s^2}=a\)

Given that Newton's second law is \(\displaystyle F=ma\), we can find the acceleration by first determining the force.

To find the gravitational force, use Newton's law of universal gravitation:

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

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

\(\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})\)

\(\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})\)

\(\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})\)

\(\displaystyle F_G=2.53*10^{-5}N\)

We now have values for both the mass and the force. Using the original equation, we can now solve for the acceleration.

\(\displaystyle F=ma\)

\(\displaystyle 2.53*10^{-5}N=(7.12*10^{18}kg)a\)

\(\displaystyle \frac{2.53*10^{-5}N}{7.12*10^{18}kg}{}=a\)

\(\displaystyle 3.55*10^{-24}\frac{m}{s^2}=a\)

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:

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

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:

\(\displaystyle F=G\frac{m_1m_2}{r^2}\) can turn into: \(\displaystyle F=m_2*\frac{G*m_1}{r^2}\) and \(\displaystyle F=m_1*\frac{G*m_2}{r^2}\), respectively.

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

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

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

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

Let's say that \(\displaystyle \frac{G*m_1}{r^2}=a\) to simplify. 

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

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

For this question, use the law of universal gravitation:

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

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.

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

\(\displaystyle F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}\)

\(\displaystyle F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})\)

\(\displaystyle F=6.37*10^{13}N\)

This force will apply to both objects in question. As it turns out, it does not matter which mass we're looking at; the force of gravity on each mass will be the same. This is supported by Newton's third law.

For this question, use the law of universal gravitation:

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

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.

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

\(\displaystyle F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}\)

\(\displaystyle F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})\)

\(\displaystyle F=6.37*10^{13}N\)

This force will apply to both objects in question. As it turns out, it does not matter which mass we're looking at; the force of gravity on each mass will be the same. This is supported by Newton's third law.

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:

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

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.

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

\(\displaystyle F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}\)

\(\displaystyle F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})\)

\(\displaystyle F=6.37*10^{13}N\)

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

\(\displaystyle F=ma\)

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

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

\(\displaystyle 0.00637\frac{m}{s^2}=a\)

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:

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

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.

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

\(\displaystyle F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}\)

\(\displaystyle F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})\)

\(\displaystyle F=6.37*10^{13}N\)

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

\(\displaystyle F=ma\)

\(\displaystyle 6.37*10^{13}N=(5.972*10^{24}kg)*a\)

\(\displaystyle \frac{6.37*10^{13}N}{5.972*10^{24}kg}{}=a\)

\(\displaystyle 1.07*10^{-11}\frac{m}{s^2}=a\)

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

\(\displaystyle F=ma\)

We know the masses, but first we need to find the forces in order to draw a conclusion about the satellites' accelerations. For this calculation, use the law of universal gravitation:

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

We can write this equation in terms of each object:

\(\displaystyle F_1=m_1*\frac{G*m_2}{r^2}\)

\(\displaystyle F_2=m_2*\frac{G*m_1}{r^2}\)

We know that the force applied to each object will be equal, so we can set these equations equal to each other.

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

We know that the second object is twice the mass of the first.

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

We can substitute for the acceleration to simplify.

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

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

The acceleration for \(\displaystyle m_1\) is twice the acceleration for \(\displaystyle m_2\); thus, the lighter mass will have the greater acceleration.

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

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

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

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

Since \(\displaystyle G\) 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.

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

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

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

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

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

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:

 \(\displaystyle \frac{1}{4}g=a\)

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