Universal Gravitation - AP Physics 1

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Question

A satellite orbits above the Earth. What is the period of the satellite's orbit?

$m_E=5.9710^{24}kg$

$G=6.6710^{-11}\frac{m^3}{kg*s}$

Answer

The period describes how long it takes the satellite to make one full orbit. If you go back to the definition of velocity, $v=\frac{\Delta x}{\Delta t}$ , 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: $v=\frac{2\pi r}{T}$ . 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 .

Solve for for the satellite. To do this, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that 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.

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

$F_G=(6.6710^{-11}\frac{m^3}{kgs})\frac{(2500kg)(5.9710^{24}kg)}{(6.3710^5m+3.810^8m)^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{1.4910^{28}kg^2}{1.4510^{17}m^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})(1.0310^{11}\frac{kg^2}{m^2})$

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

$2.710^{-3}\frac{m}{s^2}=a_c$

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!

$a_c=\frac{v^2}{r}$

$2.710^{-3}\frac{m}{s^2}=\frac{v^2}{1.4510^{17}m}$

$(2.710^{-3}\frac{m}{s^2})(1.4510^{17}m)={v^2}$

$2.8510^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.8510^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.3310^7\frac{m}{s}=v$

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.

$v=\frac{\Delta x}{\Delta t}=\frac{2\pi r}{T}$

$5.3310^7\frac{m}{s}=\frac{2\pi(1.4510^{17}m)}{T}$

$T=\frac{2\pi(1.4510^{17}m)}{5.3310^7\frac{m}{s}}$

$T=1.7110^{11}s$

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Question

A satellite orbits above the Earth. The satellite runs into another stationary satellite of equal mass and the two stick together. What is their resulting velocity?

$m_E=5.9710^{24}kg$

$G=6.6710^{-11}\frac{m^3}{kg*s}$

Answer

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

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

$(2500kgv_1)+(2500kg0\frac{m}{s})=(2500kg+2500kg)v_3$

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 .

Solve for for the satellite. To do this, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that 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.

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

$F_G=(6.6710^{-11}\frac{m^3}{kgs})\frac{(2500kg)(5.9710^{24}kg)}{(6.3710^5m+3.810^8m)^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{1.4910^{28}kg^2}{1.4510^{17}m^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})(1.0310^{11}\frac{kg^2}{m^2})$

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

$2.710^{-3}\frac{m}{s^2}=a_c$

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!

$a_c=\frac{v^2}{r}$

$2.710^{-3}\frac{m}{s^2}=\frac{v^2}{1.4510^{17}m}$

$(2.710^{-3}\frac{m}{s^2})(1.4510^{17}m)={v^2}$

$2.8510^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.8510^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.3310^7\frac{m}{s}=v$

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.

$(2500kg5.3310^{7}\frac{m}{s})+(2500kg0\frac{m}{s})=(2500kg+2500kg)v_3$

$1.332510^{11}\frac{kgm}{s}=(5000kg)v_3$

$\frac{1.332510^{11}\frac{kgm}{s}}{5000kg}=v_3$

$2.67*10^7\frac{m}{s}=v_3$

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Question

If on earth you have a weight, , what would your new weight be if you were standing on a planet with the same mass as earth, but with half the radius?

Answer

Your weight is a function of how much force is pulling you down towards the planet, not just your mass. To find force when your standing on a different planet, you would need to use Newton's Law of Universal Gravitation.

$F = - \frac{G M}{r^{2}}$

To compare the forces from each planet, you would set this equation equal to itself.

$- \frac{G M_{earth}}{r_{earth}^{2}} = - \frac{G M_{planet}}{r_{planet}^{2}}$

We then cancel the common terms, in this formula that's the negative sign, (the gravitational constant), and the mass of the earth/planet (because they're the same). After that we can substitute the radius of the new planet for half of the earth radius.

$r_{planet}=\frac{1}{2}{}r_{earth}$

We need to remember that when we square the radius in the Law of Universal Gravitation, we also need to square the $\frac{1}{2}$ , making it $\frac{1}{4}$ . Because the $\frac{1}{4}$ is in the denominator, we take the inverse of it, so you would feel times more force standing on the new planet. Therefore you would have times as much weight.

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Question

A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (ag) compare to the acceleration at the surface of Earth (g)?

Answer

The acceleration of gravity is given by the equation $a_{g} = \frac{GM}{r^{2}}$ , where G is constant.

For Earth, $a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g$ .

For the new planet,

$a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g$ .

So, the acceleration is the same in both cases.

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Question

Two satellites in space, each with a mass of , are apart from each other. What is the force of gravity between them?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

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

$F_G=G\frac{m_1m_2}{r^2}$

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

$F_G=(6.6710^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(2000kg) (2000kg)}{(1500m)^2})$

$F_G=(6.6710^{-11}\frac{m^3}{kg\cdot s^2})(\frac{4000000kg^2}{2250000m^2})$

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

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Question

Two satellites in space, each with a mass of , are apart from each other. What is the force of gravity between them?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

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

$F_G=G\frac{m_1m_2}{r^2}$

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

$F_G=(6.6710^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(1723kg )(1723kg)}{(890m)^2})$

$F_G=(6.6710^{-11}\frac{m^3}{kg\cdot s^2})(\frac{2968729kg^2}{792100m^2})$

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

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Question

Two satellites in space, each with equal mass, are apart from each other. If the force of gravity between them is $2.8810^{-10}N$ , what is the mass of each satellite?

$\small G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

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

$F_G=G\frac{m_1m_2}{r^2}$

We are given the value of the force, the distance (radius), and the gravitational constant. We are also told that the masses of the two satellites are equal. Since the masses are equal, we can reduce the numerator of the law of gravitation to a single variable.

$F_G=G\frac{M^2}{r^2}$

Now we can use our give values to solve for the mass.

$2.8810^{-10}N=(6.6710^{-11}\frac{m^3}{kg\cdot s^2})(\frac{M^2}{(80m)^2})$

$\frac{2.8810^{-10}N}{(6.6710^{-11}\frac{m^3}{kg\cdot s^2})}=\frac{M^2}{6400m^2}$

$4.32\frac{kg^2}{m^2}=\frac{M^2}{6400m^2}$

$6400m^2* 4.32\frac{kg^2}{m^2}=M^2$

$\sqrt{27648kg^2}=\sqrt{M^2}$

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Question

Two asteroids in space are in close proximity to each other. Each has a mass of $6.6910^{15}kg$ . If they are apart, what is the gravitational force between them?

$\small G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

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

$F_G=G\frac{m_1m_2}{r^2}$

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

$F_G=(6.6710^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(6.6910^{15}kg )(6.6910^{15}kg )}{(100,000m)^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}(\frac{4.4810^{31}kg^2}{1010^9m^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}(4.4810^{21}\frac{kg^2}{m^2})$

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

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Question

Two asteroids in space are in close proximity to each other. Each has a mass of $6.6910^{15}kg$ . If they are apart, what is the gravitational acceleration that they experience?

$\small G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

Given that , 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:

$F_G=G\frac{m_1m_2}{r^2}$

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

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}(\frac{(6.6910^{15}kg)(6.6910^{15}kg)}{(100,000m)^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (\frac{4.4810^{31}kg^2}{1010^9m^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}(4.4810^{21}\frac{kg^2}{m^2})$

$F_G=2.9910^{11}N$

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

$2.9910^{11}N=(6.6910^{15}kg)a$

$\frac{2.9910^{11}N}{6.6910^{15}kg}{}=a$

$4.4710^{-5}\frac{m}{s^2}=a$

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Question

Two asteroids, one with a mass of $7.1210^{18}kg$ and the other with mass , are $1010^{10}m$ apart. What is the gravitational force on the LARGER asteroid?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

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

$F_G=G\frac{m_1m_2}{r^2}$

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

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.1210^{18}kg)(5.3310^{8}kg)}{(1010^{10}m)^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.7910^{27}kg}{1010^{21}m^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (3.7910^5\frac{kg^2}{m^2})$

$F_G=2.5310^{-5}N$

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.

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Question

Two asteroids, one with a mass of $7.1210^{18}kg$ and the other with mass are $1010^{10}m$ apart. What is the gravitational force on the SMALLER asteroid?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

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

$F_G=G\frac{m_1m_2}{r^2}$

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

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.1210^{18}kg)(5.3310^{8}kg)}{(1010^{10}m)^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.7910^{27}kg}{1010^{21}m^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (3.7910^5\frac{kg^2}{m^2})$

$F_G=2.5310^{-5}N$

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.

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Question

Two asteroids, one with a mass of $7.1210^{18}kg$ and the other with mass are $1010^{10}m$ apart. What is the acceleration of the SMALLER asteroid?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

Given that Newton's second law is , we can find the acceleration by first determining the force.

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

$F_G=G\frac{m_1m_2}{r^2}$

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

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.1210^{18}kg)(5.3310^{8}kg)}{(1010^{10}m)^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.7910^{27}kg}{1010^{21}m^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (3.7910^5\frac{kg^2}{m^2})$

$F_G=2.5310^{-5}N$

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

$2.5310^{-5}N=(5.3310^8kg)a$

$\frac{2.5310^{-5}N}{5.3310^8kg}{}=a$

$4.74*10^{-14}\frac{m}{s^2}=a$

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Question

Two asteroids, one with a mass of $7.1210^{18}kg$ and the other with mass are $1010^{10}m$ apart. What is the acceleration of the LARGER asteroid?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Answer

Given that Newton's second law is , we can find the acceleration by first determining the force.

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

$F_G=G\frac{m_1m_2}{r^2}$

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

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.1210^{18}kg)(5.3310^{8}kg)}{(1010^{10}m)^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.7910^{27}kg}{1010^{21}m^2})$

$F_G=6.6710^{-11}\frac{m^3}{kg\cdot s^2} (3.7910^5\frac{kg^2}{m^2})$

$F_G=2.5310^{-5}N$

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

$2.5310^{-5}N=(7.1210^{18}kg)a$

$\frac{2.5310^{-5}N}{7.1210^{18}kg}{}=a$

$3.55*10^{-24}\frac{m}{s^2}=a$

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Question

An asteroid with a mass of $1010^{15}kg$ approaches the Earth. If they are apart, what is the gravitational force exerted by the asteroid on the Earth?

$\small M_{Earth}=5.97210^{24}kg$

$\small G=6.6710^{-11}\frac{m^3}{kgs^2}$

Answer

For this question, use the law of universal gravitation:

$F=G\frac{m_1m_2}{r^2}$

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.

$F=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{(1010^{15}kg)(5.97210^{24}kg)}{(250,000,000m)^2}$

$F=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{5.97210^{40}kg^2}{6.2510^{16}m^2}$

$F=(6.6710^{-11}\frac{m^3}{kgs^2})(9.555210^{23}\frac{kg^2}{m^2})$

$F=6.3710^{13}N$

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.

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Question

Pluto radius:

Pluto mass: $1.3*10^{22} kg$

Determine the gravity constant, $g_{pluto}$ on the surface of Pluto.

Answer

Set both forms of gravitational force equal to each other.

$m_{object}g_{pluto}=-G\frac{m_{pluto} m_{object}}{r^2}$

Simplify:

$g_{pluto}=-G\frac{m_{pluto}}{r^2}$

Plug in values:

$g_{pluto}=-6.67410^{-11}\frac{1.310^{22}}{(1.186*10^6)^2}$

$g_{pluto}=-.616\frac{m}{s^2}$

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Question

An asteroid with a mass of $1010^{15}kg$ approaches the Earth. If they are apart, what is the gravitational force exerted by the Earth on the asteroid?

$\small M_{Earth}=5.97210^{24}kg$

$\small G=6.6710^{-11}\frac{m^3}{kgs^2}$

Answer

For this question, use the law of universal gravitation:

$F=G\frac{m_1m_2}{r^2}$

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.

$F=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{(1010^{15}kg)(5.97210^{24}kg)}{(250,000,000m)^2}$

$F=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{5.97210^{40}kg^2}{6.2510^{16}m^2}$

$F=(6.6710^{-11}\frac{m^3}{kgs^2})(9.555210^{23}\frac{kg^2}{m^2})$

$F=6.3710^{13}N$

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.

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Question

An asteroid with a mass of $1010^{15}kg$ approaches the Earth. If they are apart, what is the asteroid's resultant acceleration?

$\small M_{Earth}=5.97210^{24}kg$

$\small G=6.6710^{-11}\frac{m^3}{kgs^2}$

Answer

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

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

$F=G\frac{m_1m_2}{r^2}$

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.

$F=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{(1010^{15}kg)(5.97210^{24}kg)}{(250,000,000m)^2}$

$F=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{5.97210^{40}kg^2}{6.2510^{16}m^2}$

$F=(6.6710^{-11}\frac{m^3}{kgs^2})(9.555210^{23}\frac{kg^2}{m^2})$

$F=6.3710^{13}N$

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

$6.3710^{13}N=(1010^{15}kg)a$

$\frac{6.3710^{13}N}{10*10^{15}kg}{}=a$

$0.00637\frac{m}{s^2}=a$

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Question

An asteroid with a mass of $1010^{15}kg$ approaches the Earth. If they are away, what is the Earth's resultant acceleration?

$M_{Earth}=5.97210^{24}kg$

$G=6.6710^{-11}\frac{m^3}{kgs^2}$

Answer

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

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

$F=G\frac{m_1m_2}{r^2}$

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.

$F=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{(1010^{15}kg)(5.97210^{24}kg)}{(250,000,000m)^2}$

$F=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{5.97210^{40}kg^2}{6.2510^{16}m^2}$

$F=(6.6710^{-11}\frac{m^3}{kgs^2})(9.555210^{23}\frac{kg^2}{m^2})$

$F=6.3710^{13}N$

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

$6.3710^{13}N=(5.97210^{24}kg)a$

$\frac{6.3710^{13}N}{5.97210^{24}kg}{}=a$

$1.0710^{-11}\frac{m}{s^2}=a$

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Question

Two satellites are a distance from each other in space. If one of the satellites has a mass of and the other has a mass of , which one will have the greater acceleration?

Answer

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

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:

$F=G\frac{m_1m_2}{r^2}$

We can write this equation in terms of each object:

$F_1=m_1\frac{Gm_2}{r^2}$

$F_2=m_2\frac{Gm_1}{r^2}$

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

$m_1\frac{Gm_2}{r^2}=m_2\frac{Gm_1}{r^2}$

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

$m_1\frac{G2m_1}{r^2}=2m_1\frac{Gm_1}{r^2}$

We can substitute for the acceleration to simplify.

$m_12(\frac{Gm_1}{r^2})=2m_1\frac{Gm_1}{r^2}$

The acceleration for is twice the acceleration for ; thus, the lighter mass will have the greater acceleration.

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Question

How would the linear velocity of the Moon be different if it's mass was doubled? Assume that the distance to the Earth stayed the same.

Answer

Set centripetal force equal to gravitational force:

$m_M\frac{v^2}{r}=G\frac{m_M m_E}{r^2}$

$\frac{v^2}{r}=G\frac{m_E}{r^2}$

The mass of the Moon, cancels out, thus, there is no effect.

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