Understanding Universal Gravitation

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Physics › Understanding Universal Gravitation

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Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are apart, what is the gravitational force between them?

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

$2.99*10^{11}N$

$3.01*10^{-8}N$

$5.98*10^{12}N$

$1.49*10^{-32}N$

$1.11*10^{30}N$

Explanation

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.

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

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

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

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

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass , are $10*10^{10}m$ apart. What is the gravitational force on the LARGER asteroid?

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

$2.53*10^{-5}N$

$4.74*10^{-6}N$

$3.55*10^{-6}N$

$4.61*10^{-10}N$

Explanation

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.

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

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

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

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

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.

An astronaut lands on a planet with twelve times the mass of Earth and the same radius. What will be the acceleration due to gravity on this planet, in terms of the acceleration due to gravity on Earth?

$\frac{1}{144}g$

$\frac{1}{12}g$

$g\sqrt{12}$

Explanation

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

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

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

$G\frac{m*m_{E}}{r^2}=mg$

Notice that the mass cancels out from both sides.

$G\frac{m_{E}}{r^2}=g$

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

The new planet has a mass equal to twelve times that of Earth. That means it has a mass of . It has the same radius as Earth, . Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

$G\frac{12m_{E}}{r^2}=a$

$12(G\frac{m_{E}}{r^2})=a$

We had previously solved for the gravity on Earth:

$G\frac{m_{E}}{r^2}=g$

We can substitute this into the new acceleration equation:

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

An asteroid with a mass of $10*10^{15}kg$ approaches the Earth. If they are apart, what is the gravitational force exerted by the asteroid on the Earth?

$\small M_{Earth}=5.972*10^{24}kg$

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

$6.37*10^{13}N$

$1.27*10^{14}N$

$6.7*10^{30}N$

$1.59*10^{22}N$

Explanation

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.

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

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

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

$F=6.37*10^{13}N$

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.

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?

They will have the same acceleration

The acceleration of each satellite will be zero

We need to know the value of the masses to solve

Explanation

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:

$F_1=m_1*\frac{G*m_2}{r^2}$

$F_2=m_2*\frac{G*m_1}{r^2}$

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

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

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

$m_1*\frac{G*2m_1}{r^2}=2m_1*\frac{G*m_1}{r^2}$

We can substitute for the acceleration to simplify.

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

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

Two planets are apart. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the gravitational force between them?

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

$1.24*10^{20}N$

$4.7*10^{-1}N$

$2.23*10^{18}N$

$1.24*10^{19}N$

$1.86*10^{30}N$

Explanation

To solve, 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.

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

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

A satellite orbits above the Earth. What is the tangential velocity of the satellite?

$m_E=5.97*10^{24}kg$

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

$5.33*10^7\frac{m}{s}$

$6.87\frac{m}{s}$

$1.24*10^{-31}\frac{m}{s}$

$2.85*10^{15}\frac{m}{s}$

$2.7*10^{-3}\frac{m}{s}$

Explanation

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 .

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.

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

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

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

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

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.

$2.7*10^{-3}\frac{m}{s^2}=a_c$

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!

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

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

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

$2.85*10^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.33*10^7\frac{m}{s}=v$

An astronaut lands on a planet with twice the mass of Earth, and half of the radius. What will be the acceleration due to gravity on this planet, in terms of the acceleration due to gravity on Earth?

$\frac{1}{2}g$

Explanation

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

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

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

$G\frac{m*m_{E}}{r^2}=mg$

Notice that the mass cancels out from both sides.

$G\frac{m_{E}}{r^2}=g$

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 . It also has half the radius of Earth, $\small \frac{1}{2}r$ . Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

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

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

$G\frac{2m_{E}}{\frac{1}{4}r^2}=a$

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

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

We had previously solved for the gravity on Earth:

$G\frac{m_{E}}{r^2}=g$

We can substitute this into the new acceleration equation:

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

An astronaut lands on a new planet. She knows her own mass, , and the radius of the planet, . What other value must she know in order to find the mass of the new planet?

The force of gravity she exerts on the planet

The orbit of the planet

Air pressure on the planet

The density of the planet

The planet's distance from Earth

Explanation

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

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

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

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

Since 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.

An asteroid with a mass of $10*10^{15}kg$ approaches the Earth. If they are apart, what is the gravitational force exerted by the Earth on the asteroid?

$\small M_{Earth}=5.972*10^{24}kg$

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

$6.37*10^{13}N$

$7.22*10^{14}N$

$1.59*10^{40}N$

$1.59*10^{22}N$

Explanation

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.

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