AP Physics 1 - Universal Gravitation

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.

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

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

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

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

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.

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.

Mass of Jupiter: $1.89*10^{27}kg$

Universal gravitation constant: $6.67*10^{-11}\frac{N*m^2}{kg^2}$

Radius of Jupiter:

A marble is placed from the surface of Jupiter. Determine the acceleration due to the gravity of Jupiter.

$4.52*10^{-2}\frac{m}{s^2}$

None of these

$4.52*10^{2}\frac{m}{s^2}$

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

$6.11*10^{-2}\frac{m}{s^2}$

Explanation

Using

and

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

Combining equations

$m_1a=G\frac{m_1m_2}{r^2}$

Solving for

$a=G\frac{m_2}{r^2}$

Plugging in values:

$a=6.67*10^{-11}\frac{1.89*10^{27}}{(1670*10^6)^2}$

$a=4.52*10^{-2}\frac{m}{s^2}$

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.

Radius of the moon: $1.737*10^{6}m$

Mass of moon: $7.35*10^{22}kg$

Jennifer is piloting her spaceship around the moon. How fast does she need to go to oribit the moon above the surface.

$1590\frac{m}{s}$

$2506\frac{m}{s}$

$666\frac{m}{s}$

$790\frac{m}{s}$

Explanation

The radius of the orbit will be the radius of the moon plus the altitude of the orbit.

$r_{orbit}=altitude+r_{moon}$

Converting to and plugging in values:

$r_{orbit}=2.0*10^{5}+1.737*10^{6}m$

$r_{orbit}=1.94*10^{6}m$

Centripetal force will need to equal universal gravitational force

$m_{ship}\frac{v^2}{r}=G\frac{m_{ship}m_{moon}}{r^2}$

Solving for velocity

$v=\sqrt{\frac{Gm_{moon}}{r}}$

Plugging in values:

$v=\sqrt{\frac{6.674*10^{-11}*7.35*10^{22}}{1.94*10^6}}$

$v=1590\frac{m}{s}$

Moon radius:

Moon mass: $7.34*10^{22} kg$

$G=6.674*10^{-11} \frac{N*m^2}{kg^2}$

Calculate the gravity constant for objects on the surface of the moon.

$g_{moon}=1.62\frac{m}{s^2}$

$g_{moon}=6.13\frac{m}{s^2}$

$g_{moon}=2.44\frac{m}{s^2}$

$g_{moon}=11.22\frac{m}{s^2}$

$g_{moon}=4.55\frac{m}{s^2}$

Explanation

Universal Gravitation law:

$F_{grav}=-G\frac{m_1m_2}{r^2}$

Assume is the Moon, and is the object on the surface, then:

$-G\frac{m_1}{r^2}$

Will be analogous to in $F_{grav}=mg$ , which will be a negative number since it is pointing down.

Plug in values:

$-6.674*10^{-11}\frac{7.34*10^{22}}{(1,740*10^3)^2}=g_{moon}$

$g_{moon}=1.62\frac{m}{s^2}$

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)?

$\frac{g}{9}$

$\frac{g}{3}$

Explanation

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.

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.

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