Universal Gravitation - AP Physics 1

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Question

Two planets are apart. If the first planet has a mass of $2.6510^{20}kg$ and the second has a mass of $6.3310^{26}kg$ , what is the acceleration on the smaller planet?

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

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Answer

Remember that Newton's second law states that . 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 for the force, use Newton's law of universal gravitation:

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

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.

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

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{(2.6510^{20}kg)(6.3310^{26}kg)}{(310^8m)^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{1.6774510^{47}kg^2}{910^{16}m^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})1.86410^{30}\frac{kg^2}{m^2}$

$F_G=1.2410^{20}N$

Now that we know the force of gravity, we can use Newton's second law and the mass of the smaller planet to solve for the acceleration of gravity.

$1.2410^{20}N=(2.6510^{20}kg)a$

$\frac{1.2410^{20}N}{2.6510^{20}kg}=a$

$0.47\frac{m}{s^2}=a$

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Question

Two planets are apart. If the first planet has a mass of $2.6510^{20}kg$ and the second has a mass of $6.3310^{26}kg$ , what is the acceleration on the larger planet?

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

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Answer

Remember that Newton's second law states that . 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 for the force, use Newton's law of universal gravitation:

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

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.

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

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{(2.6510^{20}kg)(6.3310^{26}kg)}{(310^8m)^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{1.6774510^{47}kg^2}{910^{16}m^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})1.86410^{30}\frac{kg^2}{m^2}$

$F_G=1.2410^{20}N$

Now that we know the force of gravity, we can use Newton's second law and the mass of the larger planet to solve for the acceleration of gravity.

$1.2410^{20}N=(6.3310^{26}kg)a$

$\frac{1.2410^{20}N}{6.3310^{26}kg}=a$

$1.96*10^{-7}\frac{m}{s^2}=a$

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Question

An astronaut has a mass of . He travels to a new planet and observes his weight is on this planet's surface. If the radius of the planet is , what is the mass of the planet?

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

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Answer

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

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

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

$F_g=mg=G\frac{m_1m_2}{r^2}$

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.

$510N=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{(71kg)m_2}{(610^7m)^2}$

$\frac{510N}{6.6710^{-11}\frac{m^3}{kgs^2}}=\frac{(71kg)m_2}{(610^7m)^2}$

$7.6510^{12}\frac{kg^2}{m^2}=\frac{(71kg)m_2}{3.610^{15}m^2}$

$(7.6510^{12}\frac{kg^2}{m^2})(3.610^{15}m^2)=(71kg)m_2$

$2.7510^{28}kg^2=(71kg)m_2$

$3.87*10^{26}=m_2$

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Question

There are two isolated stars orbiting each other. The first star has a mass of $2\times10^{12}kg$ and the second star has a mass of $4.5\times10^{14}kg$ . If the stars are 2,000km away, what is the gravitational force felt by the first star?

$G=6.67\times10^{-11}\frac{m^3}{kg\cdot s}$

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Answer

We need to know Newton's law of universal gravitation to solve this problem:

$F = G \frac{m_1\cdot m_2}{r^2}$

It is important to note that both suns will feel the same gravitational force. However, since they have different masses, they will accelerate at different rates.

Plugging in the variables we have, we get:

$F = 6.67\mathrm{x}10^{-11} \frac{2\mathrm{x}10^{12}kg \cdot 4.5\mathrm{x}10^{14}kg}{(2\mathrm{x}10^6m)^2}$

$F = 1.50\mathrm{x}10^4N$

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Question

A new planet is discovered with mass $3.21\cdot 10^{23}kg$ and with a diameter of $8.53\cdot 10^{3}km$ . What is the lowest escape velocity required to escape this planet's gravitational pull?

$G=6.67\cdot10^{-11}\frac{N\cdot m^2}{kg^2}$

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Answer

The equation to calculate the escape velocity from a planet is

$v_{e}=\sqrt{\frac{2GM_{p}}{R_{p}}}$

The diameter of the planet is given and can be divided by two to determine the radius of the planet. By plugging in the given values, the escape velocity threshold can be determined:

$v_{e}=\sqrt{\frac{2(6.67\cdot10^{-11}\frac{N\cdot m^2}{kg^2})(3.21\cdot10^{23}kg)}{\frac{(8.53\cdot10^{6}m)}{}2}}=3169\frac{m}{s}$

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Question

Suppose that a person on Earth weighs 800N. If this person were to travel to a distant planet that had twice the density and the same radius of Earth, how much will the person weigh on this new planet?

$G=6.67384*10^{-11} \frac{m^{3}}{kg\cdot s^{2}}$

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Answer

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:

$F=G\frac{m_{1}m_{2}}{r^{2}}$

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

$m_{1}$ is the mass of object 1

$m_{2}$ is the mass of object 2

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:

$F_{Earth}=G\frac{m_{person}m_{Earth}}{r^{2}}=800N$

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:

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

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Question

For a planet of mass $2.34*10^{22}kg$ and diameter what is the force of gravity on an object of mass ?

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Answer

To solve this we use the universal gravitation formula

$\frac{Gm_{1}m_{2}}{r^{2}}$

where G is the gravitational constant $6.67408*10^{-11} m^3 kg^{-1} s^{-2}$

and r is the radius of the planet $\frac{12000}{2}=6000km$

plugging everything in we get $F_{g}=.52 N$

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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?

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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)?

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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?

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

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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?

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

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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?

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

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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?

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

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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?

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

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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?

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

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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?

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

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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?

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

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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?

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

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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?

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

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

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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.

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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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