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In this Physics tutorial, you will learn:

- Express the gravitational potential energy in terms of masses of objects and the distance between them
- Know the meaning of cosmic velocities and what they are used for
- Identify the three Kepler Laws
- Know how to relate Kepler Laws to rotational motion equations
- Know how to use Kepler Laws to study the Universe
- Know the meaning and formula of gravitational potential
- Know how gravitational potential is related to gravitational potential energy

Do you think celestial bodies move randomly or they have a certain regularity when they move in the sky?

Why planets do not collapse with each other and with the Sun?

Is it possible to calculate the mass of celestial bodies from the Earth?

Do planets rotate at the same speed around the Sun?

We will provide answers for the above and other questions during the explanation of this topic. Therefore, please read the tutorial carefully.

In the Physics tutorial "Gravitational Potential Energy. The Law of Mechanical Energy Conservation", we explained that gravitational potential energy as a special case of potential energy is obtained by multiplying the gravitational force F_{g} exerted on a system and linear distance r the object moves due to this force, i.e.

PE = -F*⃗* × r*⃗*

in general, and

GPE = -F*⃗*_{g} × r*⃗*

in the specific case of gravitational potential energy.

Thus, since when multiplying in scalar mode a force and linear distance the object on which this force acts moves, we obtain the work done on this object, we obtain the definition of potential energy.

"**Potential energy PE represents the work done by one of the objects in the system (usually the largest) to bring the other object from the position r ⃗ to zero, i.e. to bring it at the place where the first object is**."

The **negative sign** provides the convention that work done against a force field increases the **potential energy**, while work done by the force field decreases the **potential energy**.

In the specific case of gravitational potential energy, since gravitational force is

F_{g} = G × *M × m**/**r*^{2}

where M is the mass of the largest object and m is that of the smallest object, we obtain for the gravitational potential energy possessed by an object when it is at a linear distance R from the Earth

GPE = -F*⃗*_{g} × r*⃗*

= -G ×*M × m**/**r*^{2} × r*⃗*

= -G ×*M × m**/**r*

= -G ×

= -G ×

**Remark!** The simplified equation for gravitational potential energy

GPE = m × g × h

can be used only in very specific situations in which the magnitude of gravitational field g is constant, i.e. when objects are very close to Earth surface (for small heights, h). It is known that gravitational field weakens when moving away from the source (here the Earth). Therefore, the only formula we can use to calculate the gravitational field at large distances from the source, is

GPE = -G × *M × m**/**r*

What is the work done by the Earth to move a 4 kg object from h = 29 km above its surface to the ground? Take M_{Earth} = 5.972 × 1024 kg, G = 6.674 × 10-11 N × m^{2} / kg^{2} and R_{Earth} = 6371 km.

We must calculate the gravitational potential energy at the beginning and the end of process and subtract them in order to calculate the work done by the Earth to send the 4 kg object from h = 29 km to the ground (h = 0).

Clues:

r_{1} = R + h = 6371 km + 29 km = 6400 km = 6 400 000 m = 6.4 × 10^{6} m

r_{2} = R = 6371 km = 6 371 000 m = 6.371 × 10^{6} m

M_{Earth} = M = 5.972 × 10^{24} kg

m_{object} = m = 4 kg

W = ?

r

M

m

W = ?

We have:

GPE_{1} = -G × *M × m**/**r*_{1}

= -6.674 × 10^{-11} × *5.972 × 10*^{24} × 4*/**6.4 × 10*^{6}

= -2.4910705 × 10^{8} J

= -6.674 × 10

= -2.4910705 × 10

and

GPE_{2} = -G × *M × m**/**r*_{2}

= -6.674 × 10^{-11} × *5.972 × 10*^{2}4 × 4*/**6.371 × 10*^{6}

= -2.5024095 × 10^{8} J

= -6.674 × 10

= -2.5024095 × 10

Thus, the work done by Earth's gravity on the object is

W = ∆GPE = GPE_{1} - GPE_{2}

= -2.4910705 × 10^{8} J - (-2.5024095 × 10^{8} J)

= 0.011339 × 10^{8} J

= 1.1339 × 10^{6} J

= -2.4910705 × 10

= 0.011339 × 10

= 1.1339 × 10

Thus, we obtained a result equal to 1 133 900 Joules approximately.

Now, observe what result we would obtain if we used the simplified equation GPE = m × g × h for gravitational potential energy. If we take m = 4 kg, g = 9.81 m/s^{2} and h = 29 km = 29000 m, the result is

W = ∆GPE

= m × g × h

= 4 × 9.81 × 29000

= 1 137 960 J

= m × g × h

= 4 × 9.81 × 29000

= 1 137 960 J

The last result is slightly bigger that the result obtained earlier because of the error made when considering the gravity g as constant (9.81 m/s^{2}). It is a known fact that gravity decreases with the increase in altitude. It has a value of 9.81 m/s^{2} only near the Earth surface. Therefore, in exercises it is better to use as much as possible the general formula of gravitational potential energy

GPE = - G × *M × m**/**r*

instead of the simplified formula

GPE = m × g × h

In the tutorial "Gravitational Potential Energy. The Law of Mechanical Energy Conservation", it was also explained the concept of **path independence** of gravitational potential energy. This property derives from path independence of gravitational force. This means the path is not important for the values of gravitational force and gravitational potential energy but only the initial and final positions of the object.

When you throw an object upwards, it eventually falls on the ground after a while, due to the attraction effect of Earth's gravity. But is it possible to throw an object upwards with such a velocity as to overcome the gravity?

The answer is: Yes! The condition for this, is that the force used to move the object must be enough to overcome the gravitational force and thus, turn it into centripetal (or centrifugal if you wish) force, which enables the object rotate around the Earth without falling on it. No matter whether the object rotates or not; such an action is provided by the Earth during its daily rotation around itself.

Thus, we have:

F_{g} = F_{C}

G ×*M × m**/**R*^{2}* = **m × v*^{2}*/**R*

G ×

where M is the mass of Earth, m is the mass of the object candidate to become a satellite and R, the radius of Earth. Therefore, we obtain after simplifications,

v = √

As you see, all the above values are constants. This means the throwing velocity required to turn an object into a satellite is independent from the object features. Thus, substituting the known values, we obtain

v = √*6.674 × 10*^{-11} × 5.972 × 10^{2}4*/**6.371 × 10*^{6}

= √**6.256 × 10**^{7}

= 7.9 × 10^{3} *m**/**s*

= 7.9*km**/**s*

= √

= 7.9 × 10

= 7.9

The above value is known as the **first cosmic velocity** or **orbital velocity**, i.e. the least launching velocity required to make an object rotate around the Earth without falling on it. Look at the figure.

If we are not interested to keep an object rotating around the Earth as a satellite but to send it away in space, the minimum launching velocity for this action is known as **escape velocity from Earth**. It is also known as the second cosmic velocity. Let's see how we can find its value.

If we want to make the object get unaffected from the Earth's gravity in any way, we must send it away so that the initial gravitational potential energy of object is entirely cancelled by the initial kinetic energy. Thus, from the law of mechanical energy conservation (read the tutorial on "Gravitational Potential Energy. The Law of Mechanical Energy Conservation"), we have:

GPE = KE

If we obtain only the magnitudes (without the sign minus before GPE), we obtain:

G × *M × m**/**R* = *m × v*^{2}*/**2*

where M is the mass of Earth, m is the mass of the object to be sent in space and R, the radius of Earth. Therefore, we obtain after simplifications,

G × *M**/**R* = *v*^{2}*/**2*

v^{2} = *2 × G × M**/**R*

v = √*2 × G × M**/**R*

v

v = √

Substituting the known values, we obtain for the escape velocity (the second cosmic velocity):

v = √*2 × G × M**/**R*

= √*2 × 6.674 × 10*^{-11} × 5.972 × 10^{24}*/**6.371 × 10*^{6}

= √**1.2512 × 10**^{8}

= 11185*m**/**s*

≈11.2*km**/**s*

= √

= √

= 11185

≈11.2

Now, let's consider the hypothetical scenario in which we want to launch an object from Earth with such a velocity that it leaves the Solar System and moves freely in deep space. This is known as the **third cosmic velocity**, or the escape velocity from the solar system. The reasoning is similar as in the previous case except that we must use the mass of Sun instead of that of Earth and the distance Sun-Earth instead of radius of Earth when making the calculations. Thus, we have:

GPE = KE

G ×*M*_{Sun} × m*/**R* = *m × v*^{2}*/**2*

G ×

Substituting the known values, (M_{Sun} = 1.989 × 1030 kg and R_{Sun-Earth} = 151.2 million km = 151.2 × 106 km = 1.512 × 108 km = 1.512 × 1011 m) and simplifying, we obtain for the minimum escape velocity from our solar system:

v = √*2 × G × M*_{Sun}*/**R*_{Sun-Earth}

= √*2 × 6.674 × 10*^{-11} × 1.989 × 10^{3}0*/**1.512 × 10*^{11}

= √**17.559 × 10**^{8}

= 4.19 × 10^{4} m/s

≈ 41.9 km/s

= √

= √

= 4.19 × 10

≈ 41.9 km/s

Due to many gravitational forces acting on the same celestial body in all directions, the trajectory is not as simple as one may think. Therefore, a complex theoretical approach must be used to study the movement of celestial bodies. Johannes Kepler was the first who formulated a scientific-based theory to explain the planetary motion. This theory is based on three fundamental laws, known as Kepler Laws. Let's explain them.

This law is otherwise known as the **Law of Orbits**. It states that:

**All planets move in elliptical (not circular) orbits, where the Sun is at one of the ellipse foci**.

This means the Sun has not always the same distance from Earth during the year. Look at the figure in which there is an upper view of Earth's orbit around the Sun:

Obviously, the elliptic trajectory shown above is exaggerated for demonstration purpose. In reality, it looks more like a circle than an ellipse. That's why we often depict planetary orbits as circular.

The shortest distance of a planet from the Sun is known as "**perihelion**" and the longest distance from the Sun as "**aphelion**".

The equations of perihelion and aphelion are as follows:

R_{p} = a × (1 - e)

and

R_{a} = a × (1 + e)

Where a is the long half-axis of ellipse and e is its eccentricity (e = c / a where c is the distance from centre of ellipse to any focus). Look at the figure:

This law derives from the principle of conservation of angular momentum described in the tutorial "Dynamics of Rotational Motion". It states that:

**Any line that connects a planet to the Sun "wipes out" equal surface areas in equal time intervals**.

This means that when a planet is closer to the Sun, it moves faster than when it is farther for it. Look at the figure.

The coloured areas are equal, so the planet in the figure moves faster from position I to II than from position III to IV for the same time interval Δt. Thus, in the above figure, we have A_{1} = A_{2}.

The Second Kepler Law is also known as the "**Law of Areas**".

This law is also known as the "**Law of Periods**". It states that:

**The square of period of a planet revolution around the Sun is proportional to the cube of the greater semi axis**.

This means that T^{2} ~ a^{3}. More precisely, we have:

T^{2} = *4 × π*^{2}*/**G × M* × a^{3}

where G is the gravitational constant and M is the mass of the Sun.

**Proof**. Starting from the known equation

F_{g} = F_{cf}

or

G *M × m**/**R*^{2} = *m × v*^{2}*/**R*

*G × M**/**R* = v^{2}

In circular motion, we have v = ω × R. Thus,

ω

Giving that

ω = 2πf = *2π**/**T*

we obtain

T

Since elliptical orbits of planets are very close to circular, we can replace R with the major semiaxis of ellipse a. Thus, we obtain the mathematical expression of 3rd Kepler Law.

T^{2} = *4π*^{2}*/**G × M* × a^{3}

This law is written in the above way when considering the orbit of planets around the Sun. Thus, since the Sun is dominant in mass, we neglect the mass of planets, but if we want to deal with orbits of Moons around their planets, we must write the third Kepler law as

T^{2} = *4 × π*^{2}*/**G × (M*_{1} + M_{2}) × a^{3}

where M_{1} is the mass of planet and M_{2} that of its Moon.

Calculate the mass of our Moon using the Kepler Laws if it completes a revolution around the Earth in 27.3 days. Take the distance from Earth to Moon equal to 384400 km and mass of Earth 5.972 × 1024 kg.

First, let's convert all values in basic units written in standard form. Given that the planetary orbits are almost circular, we can use the approximation a ≈ R. Thus, we have:

T = 27.3 d = 27.3 d × 24 h/d × 60 min/h × 60 s/min = 2 358 720 s = 2.35872 × 106 s

a ≈ R = 384 400 km = 384 400 000 m = 3.844 × 108 m

M_{Earth} = M1 = 5.972 × 1024 kg

a ≈ R = 384 400 km = 384 400 000 m = 3.844 × 108 m

M

Therefore, using the Third Kepler Law (the Law of Periods), we obtain for mass of the Moon M_{2},

T^{2} = *4 × π*^{2}*/**G × (M*_{1} + M_{2}) × a^{3}

M_{1} + M_{2} = *4 × π*^{2}*/**G × T*^{2} × a^{3}

M_{2} = *4 × π*^{2}*/**G × T*^{2} × a^{3}-M_{1}

M

M

= *4 × 3.14*^{2} × (3.844 × 10^{8})^{3}*/**6.674 × 10*^{-11} × (2.35872 × 10^{6} )^{2} - 5.972 × 10^{24}

= *39.4384 × 56.8002 × 10*^{2}4*/**6.674 × 10*^{-11} × 5.56356 × 10^{12} - 5.972 × 10^{24}

= *2240.109 × 10*^{2}4*/**37.1312 × 10*^{1} - 5.972 × 10^{24}

= 60.329 × 10^{23} - 5.972 × 10^{24}

= 6.0329 × 10^{24} - 5.972 × 10^{24}

= 0.0609 × 10^{24} kg

= 6.09 × 10^{22} kg

= 6.0329 × 10

= 0.0609 × 10

= 6.09 × 10

This result is very close to that obtained in the question 1 of the previous tutorial (5.974 × 1022 kg). However, recent measurements that are more accurate give for mass of the Moon the value 7.34767309 × 1022 kg. This difference occurs for two reasons:

- Kepler Laws do not consider the presence of other celestial bodies around, which affect the result, and
- During a complete revolution of Moon around the Earth, the orbit is neither circular, nor elliptic as the Earth itself moves around the Sun. Thus, the Moon's trajectory is more or less like this (for convenience, the Earth's orbit around the Sun is taken as circular):

The quantity

ϕ = -*G × M**/**R*

is known as "**gravitational potential**". It represents the attracting ability of a celestial body. Since G and M are constants, this attracting ability depends only on the distance from the planet. For example, gravitational potential of Earth on its surface is

ϕ = - *6.674 × 10*^{-11} *N × m*^{2}*/**v* × 5.972 × 10^{24} kg*/**6.371 × 10*^{6} m

= -6.256 × 10^{7} m^{2}/s^{2}

= -6.256 × 10

As you can see, the unit of gravitational potential is [m^{2} / s^{2}], i.e. that of square of the velocity. Indeed, gravitational potential represents the square of the first cosmic velocity (orbital velocity). The sign minus is because of the relationship between gravitational potential ϕ and gravitational potential energy GPE (or U). It is

GPE = m × ϕ

where m is mass of the object at the position R inside the gravitational field which is affected by the gravitational potential ϕ.

**Gravitational Potential Energy GPE (or U) represents the work done by gravitational field of one of the objects in the system (usually the largest) to bring the other object from the position r ⃗ to zero, i.e. to bring it at the place where the first object is**.

The equation of gravitational potential energy is

GPE = -G × *M × m**/**r*

where M is the mass of the largest object and m is that of the smallest object, we obtain for the gravitational potential energy possessed by an object when it is at a linear distance R from the Earth.

Path independence principle of gravitational field states means the path is not important for the values of gravitational force and gravitational potential energy but only the initial and final positions of the object.

The **first cosmic velocity** or **orbital velocity** represents the least velocity required to make an object rotate around the Earth without falling on it. Its equation is

v = √*G × M**/**R*

and for objects thrown from Earth surface it is about 7.9 km/s.

The **second cosmic velocity** or the **escape velocity from Earth** represents the minimum velocity to send an object away in space but still inside our solar system. Its equation is

v = √*2 × G × M**/**R*

and for objects thrown from Earth surface it is about 11.2 km/s.

If we want to launch an object from Earth with such a velocity that it leaves the Solar System and moves freely in deep space, we must apply the **third cosmic velocity**, or the escape velocity from the solar system. Its equation is

v = √*2 × G × M*_{Sun}*/**R*_{Sun-Earth}

and for objects thrown from Earth surface it is about 41.9 km/s.

Johannes Kepler was the first who formulated a scientific-based theory to explain the planetary motion. This theory is based on three fundamental laws, known as Kepler Laws. They are:

**a. First Kepler Law**

This law is otherwise known as the **Law of Orbits**. It states that:

**All planets move in elliptical (not circular) orbits, where the Sun is at one of the ellipse foci**.

**b. Second Kepler Law**

This law is known as the **Law of Areas** and it derives from the principle of conservation of angular momentum. It states that:

**Any line that connects a planet to the Sun "wipes out" equal surface areas in equal time intervals**.

**c. Third Kepler Law**

This law is also known as the "**Law of Periods**". It states that:

**The square of period of a planet revolution around the Sun is proportional to the cube of the greater semi axis**.

This means that T^{2} ~ a^{3}. More precisely, we have:

T^{2} = *4 × π*^{2}*/**G × M* × a^{3}

where G is the gravitational constant and M is the mass of the Sun.

The quantity

ϕ = -*G × M**/**R*

is known as "**gravitational potential**". It represents the attracting ability of a celestial body. Since G and M are constants, this attracting ability depends only on the distance from the planet. Gravitational potential has the unit of square of velocity, [m^{2}/s^{2}].

**1)** What is the minimum launching velocity (in m/s) a rocket must have to become a Moon's artificial satellite if it is launched from Moon's surface? Take G = 6.674 × 10-11 N × m2/ kg2, M_{Moon} = 7.348 × 1022 kg and R_{Moon} = 1737 km.

- 282 m/a
- 5313 m/s
- 2823 m/s
- 1680 m/s

**Correct Answer: D**

**2)** An object is launched at 8 km/s from Mars surface. What happens to this object after its launch?

(Mass of Mars is 6.41 × 1023 kg, radius of Mars is 3389.5 km and its distance from the Sun is 225 million km. Also take mass of the Sun equal to 1.989 × 1030 kg.)

- It falls again on Mars surface
- It turns into an artificial satellite of Mars
- It moves like a planet around the Sun
- It leaves our solar system and goes into deep space

**Correct Answer: C**

**3)** The period of rotation around the Earth of a 400 t asteroid is 2 days. How far is it from Earth's centre? Take M_{Earth} = 5.9724 × 1024 kg.

**Hint!** Consider the fact that M_{Earth} >> m_{asteroid}

- 98 222 km
- 947 617 km
- 9476 km
- 9822 km

**Correct Answer: A**

We hope you found this Physics tutorial "Gravitational Potential Energy. Kepler Laws" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In the next Physics section, we expand your insight and knowledge of "potential and potential energy" with our Physics tutorials on Electrostatics.

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