# Gravitation, orbit, escape velocity

##### Examples

###### Lessons

**Satellite in geostationary orbit of the Earth**A 465 kg satellite is in geostationary orbit around the Earth.

- What is the radius of its circular orbit? What is its altitude?
- What is the satellite's speed?
- What is the gravitational potential energy of the satellite when it is still attached to its launch rocket at the Earth's surface? What is the potential energy in orbit?
- Concept: what is the difference between $E_{p}=mgh$ and $E_{p}=-G\frac{m_{1}m_{2}}{r}$? Why is one positive and one negative?
- What is the total mechanical energy of the satellite once it is in orbit?
- Concept: which of the following sets of energies could represent the kinetic energy, gravitational potential energy, and total mechanical energy of a satellite in a stable circular orbit? (Proof: for an object in stable circular orbit, $E_{p}=-2E_{k}$)
a) $E_{k} = 30000J, E_{p} = -30000J, E_{mech} = 0J$

b) $E_{k} = 30000J, E_{p} = -60000J, E_{mech} = -30000J$

c) $E_{k} = 60000J, E_{p} = -30000J, E_{mech} = 30000J$

d) $E_{k} = 60000J, E_{p} = 30000J, E_{mech} = 90000J$

- How much work must be done on the satellite to launch it into geostationary orbit?
- What speed does the satellite need at the surface of the Earth from its launch to reach the geostationary orbit?
- In orbit, how much additional energy does the satellite need to escape the Earth's gravity?
- What is the escape velocity from the surface of the Earth?

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###### Topic Notes

In this lesson, we will learn:

- Law of universal gravitation
- Solving problems involving gravitation, orbit, and escape velocity

__Notes:__- The force of gravity between any two objects in the universe can be calculated with the
__universal law of gravitation__. The magnitude of the force of gravity is proportional to the masses of the two objects (gravity increases when larger masses are involved). Gravity is inversely proportional to the square of distance between the two centers of gravity of the two objects (this distance is called__radius__: gravity decreases as the radius gets larger). G is an experimentally determined constant that relates mass and radius to gravitational force. - Why don't we always just use $F_{g}=mg$?
- To use $F_{g}=mg$ (which is a form of Newton's second law, $F=ma$) we need to know the value of $g$, the acceleration due to gravity. $g$ depends on the distance between the two objects: if we look at the universal gravitation formula, we can see that the force of gravity decreases as objects are moved apart (increasing $r$ decreases $F_{g}$). If $F_{g}$ decreases $g$ must also decrease. For most physics problems on Earth, $g$ = 9.8 $m/s^{2}$ downwards is a good estimate even though $g$ decreases as an object moves away from the Earth's surface: because the radius of the Earth from center to surface is already so large, you must move an object many kilometers away from the surface before the magnitude of $g$ decreases significantly from 9.8 $m/s^{2}$.
- If $g$ is unknown (either because an object is moved a large distance away from the Earth, or because you want to calculate the force of gravity between two objects where neither is the Earth), then you must use the universal law of gravitation.

- We can also calculate an object's gravitational potential energy with the gravitational potential energy formula given below. Remember, this energy represents how much energy is stored in an object based on its position in a gravitational field: the further it can fall, the more energy is stored.
- To use the form of the potential energy equation that we have already used in this course ($E_{p}=mgh$) we have to know the value of $g$ and that $g$ is constant in the region that the object is moving. Just like with $F_{g}=mg$, use $E_{p}=mgh$ is useful for situations close to the Earth's surface, where we can approximate that $g$ is always 9.8 $m/s^{2}$ downwards. Use the gravitational potential energy formula for values of $g$ that are unknown, or change significantly because of changing distance between objects (such as when an object is moved a large distance from Earth's surface).

- When gravity acts between two objects, each object feels an equal force of gravity pulling it towards the other object. In our falling rock example, the rock has $F_{g (Earth \,on\, rock)}$ acting on it which causes it to accelerate down towards the Earth. Less obvious it that the Earth is also experiencing equal magnitude force $F_{g (rock\, on\, Earth)}$ which accelerates the Earth upwards towards the rock. This force is so small compared to the Earth's mass, however, that the acceleration of the Earth is essentially zero.
- Recall Newton's third law: for every action force there is an equal and opposite reaction force. If the Earth pulls on the rock with $F_{g (Earth\, on\, rock)}$, the rock must pull on the Earth with $F_{g (rock\, on\, Earth)}$ make up the force pair.

__Orbit__is when an object moves in a circular path around another object due to the force of gravity.- Orbit is a form of
__circular motion__: all of the rules, equations, and observations we have previously learned for circular motion will apply to orbit problems. Some important points to review:- For an object that is in uniform circular motion, the sum of all forces acting on the object must add up to a net force that is a
__centripetal force__: a force that is always pointed towards the center of the object's circular path.- Usually in orbit the only force on the object is the force of gravity, which can act as a centripetal force since it always points towards the center of the planet/large mass that is being orbited.

- This force produces an acceleration that always points inwards towards the center of the circular path, called
__centripetal acceleration__. This acceleration is constantly changing the direction of the object's velocity to keep it moving in a circular path. The speed of the object does not change, only direction.

- For an object that is in uniform circular motion, the sum of all forces acting on the object must add up to a net force that is a

- Orbit is a form of
__Escape velocity__is the minimum velocity needed by an object travelling away from a planet to escape its gravitational influence. An object launched from Earth at a velocity equal or greater than the Earth's escape velocity can continue to travel through space away from Earth forever. An object launched below escape velocity will either be pulled back and crash at the Earth's surface or fall into a stable orbit around the Earth.- Gravitational potential energy is defined as being zero when objects are moved so far apart that they no longer are pulled together by gravity, which would be when they are moved an "infinite distance" apart (to understand why it is defined this way, think about what happens to the equation $E_{p} = -G\frac{m_{1}m_{2}}{r}$ as $r$ is increased to become very large so that $r$ gets closer to infinity: $E_{p}$ must decrease until it is eventually zero).
- If an object launched from the Earth's surface has enough kinetic energy so that the total mechanical energy of the object is zero ($E_{p}+E_{k}=0J$), that means that the object has the minimum amount of kinetic energy needed to escape the Earth's gravity. The escape velocity is the velocity that gives this kinetic energy.
- If we imagine launching a rocket from Earth, it can be helpful to think of the negative sign on $E_{p}$ as indicating that the $E_{p}$ is "binding" the rocket to the Earth. The positive sign of the kinetic energy of the rocket indicates that it is acting to "take away" the rocket from the Earth. This gives us the following possibilities for the rocket launch:
- We launch the rocket with a kinetic energy of smaller magnitude than the potential energy, so that kinetic + potential energy = a negative energy value. The negative energy means that the rocket is still bound to the Earth: it either falls back to Earth or into orbit.
- We launch the rocket with a kinetic energy of magnitude equal to the potential energy so that kinetic energy + potential energy = zero. Since the energy is not negative, we have given the rocket the minimum amount of kinetic energy it needs to escape from the potential energy that was binding it to the Earth.
- We launch the rocket with a kinetic energy of magnitude greater than the potential energy so that kinetic energy + potential energy = a positive energy value. This means we have given the rocket more than enough energy to escape the Earth's gravity.

- If we imagine launching a rocket from Earth, it can be helpful to think of the negative sign on $E_{p}$ as indicating that the $E_{p}$ is "binding" the rocket to the Earth. The positive sign of the kinetic energy of the rocket indicates that it is acting to "take away" the rocket from the Earth. This gives us the following possibilities for the rocket launch:

**Gravitation:**

**Law of Universal Gravitation**

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

$F_{g}$: force of gravity, in newtons (N)

$G:$ gravitational constant, $6.67\times 10^{-11}$ N·$m^{2}/kg^{2}$

$m_{1}, m_{2}:$ mass of objects, in kilograms (kg)

$r:$ distance between two objects, in meters (m)

**Gravitational Potential Energy**

$E_{p} = -G \frac{m_{1}m_{2}}{r}$

$E_{p}:$ gravitational potential energy, in joules (J)

$G:$ gravitational constant, $6.67\times 10^{-11}$ N·$m^{2}/kg^{2}$

$m_{1}, m_{2}:$ mass of objects, in kilograms (kg)

$r:$ distance between two objects, in meters (m)

**Minimum Conditions for Escape from Gravity**

$E_{k}+E_{p}=0J$

$E_{k}:$ kinetic energy, in joules (J)

$E_{p}:$ gravitational potential energy, in joules (J)

**Useful Constants**

$G:$ gravitational constant, $6.67\times 10^{-11}$ N·$m^{2}/kg^{2}$

$m_{earth}: 5.98\times 10^{24} kg$

$m_{moon}: 7.35\times 10^{22} kg$

$m_{sun}: 1.98\times 10^{30} kg$

$r_{earth}: 6.38\times 10^{6} kg$

$r_{moon}: 1.74\times 10^{6} kg$

$r_{sun}: 6.96\times 10^{8} kg$

$d_{earth-moon}: 3.84\times 10^{8} m$

$d_{sun-earth}: 1.50\times 10^{11} m$

$T_{earth's rotation}: 8.61\times 10^{4} s$

**Circular Motion: **

**Period and Frequency**

$T = \frac{total time}{\# of revolutions}$

$f = \frac{\# of revolutions}{total time}$

$T = \frac{1}{f}$

$T:$ period, in seconds (s)

$f:$ frequency, in hertz (Hz)

**Centripetal Acceleration**

$a_{c} = \frac{v^{2}}{r} = \frac{4\pi^{2}r}{T^{2}}$

$a_{c}:$ centripetal acceleration, in meters per second squared $(m/s^{2})$

$v:$ velocity, in meters per second (m/s)

$r:$ radius, in meters (m)

$T:$ period, in seconds (s)

**Energy:**

**Work**

$W = F_\parallel d = \Delta E_{mech} = (E_{kf} + E_{pf}) - (E_{ki} + E_{pi})$

$W:$ work, in joules (J)

$d:$ displacement, in meters (m)

$F_\parallel:$ component of force parallel to $d$ in newtons (N)

$\Delta E_{mech}:$ change in mechanical energy

$(E_{kf} + E_{pf}):$ total final mechanical energy, in joules (J)

$(E_{ki} + E_{pi}):$ total initial of force parallel to $d$, in newtons (N)

**Kinetic Energy**

$E_k = \frac{1}{2}mv^2$

$E_k;$ kinetic energy, in joules (J)

$m:$ mass, in kilgrams (kg)

$v:$ velocity, in meters per second (m/s)

**Gravitational Potential Energy**

$E_p = mgh$

$E_p:$ gravitational potential energy, in joules (J)

$m$: mass, in kilograms (kg)

$g:$ acceleration due to gravity, in meters per second squared (m/s^{2})

$h:$ height, in meters (m)

**Conservation of Energy**

$\sum E_i = \sum E_f$

$\sum E_i:$ sum of all initial energy, in joules (J)

$\sum E_f:$ sum of all final energy, in joules (J)

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