To solve this problem, 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 satellite masses and distance (radius). Using these values we can solve for the force.

$\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(2000kg) (2000kg)}{(1500m)^2})$

$\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{4000000kg^2}{2250000m^2})$

$\displaystyle F_G=1.19*10^{-10}N$

To solve this problem, 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 satellite masses and distance (radius). Using these values we can solve for the force.

$\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(1723kg )(1723kg)}{(890m)^2})$

$\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{2968729kg^2}{792100m^2})$

$\displaystyle F_G=2.5*10^{-10}N$

To solve this problem, 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.

$\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(6.69*10^{15}kg )(6.69*10^{15}kg )}{(100,000m)^2})$

$\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$\displaystyle F_G=2.99*10^{11}N$

Given that $\displaystyle F=ma$, we already know the mass, but we need to find the force in order to solve for the acceleration.

To solve this problem, 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 satellite masses and distance (radius). Using these values we can solve for the force.

$\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{(6.69*10^{15}kg)(6.69*10^{15}kg)}{(100,000m)^2})$

$\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$\displaystyle F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$\displaystyle F_G=2.99*10^{11}N$

Now we have values for both the mass and the force, allowing us to solve for the acceleration.

$\displaystyle F=ma$

$\displaystyle 2.99*10^{11}N=(6.69*10^{15}kg)a$

$\displaystyle \frac{2.99*10^{11}N}{6.69*10^{15}kg}{}=a$

$\displaystyle 4.47*10^{-5}\frac{m}{s^2}=a$

To solve this problem, 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.

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

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

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

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

It actually doesn't matter which asteroid we're looking at; the gravitational force will be the same. This makes sense because Newton's 3rd law states that the force one asteroid exerts on the other is equal in magnitude, but opposite in direction, to the force the other asteroid exerts on it.

To solve this problem, 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.

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

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

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

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

It actually doesn't matter which asteroid we're looking at; the gravitational force will be the same. This makes sense because Newton's 3rd law states that the force one asteroid exerts on the other is equal in magnitude, but opposite in direction, to the force the other asteroid exerts on it.

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.