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|8.2||Gravitational Potential Energy. Kepler Laws|
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 Fg exerted on a system and linear distance r the object moves due to this force, i.e.
in general, and
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 r⃗ 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
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
Remark! The simplified equation for gravitational potential energy
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
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 MEarth = 5.972 × 1024 kg, G = 6.674 × 10-11 N × m2 / kg2 and REarth = 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).
Thus, the work done by Earth's gravity on the object is
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/s2 and h = 29 km = 29000 m, the result is
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/s2). It is a known fact that gravity decreases with the increase in altitude. It has a value of 9.81 m/s2 only near the Earth surface. Therefore, in exercises it is better to use as much as possible the general formula of gravitational potential energy
instead of the simplified formula
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:
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,
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
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:
If we obtain only the magnitudes (without the sign minus before GPE), we obtain:
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,
Substituting the known values, we obtain for the escape velocity (the second cosmic velocity):
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:
Substituting the known values, (MSun = 1.989 × 1030 kg and RSun-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:
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:
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 A1 = A2.
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 T2 ~ a3. More precisely, we have:
where G is the gravitational constant and M is the mass of the Sun.
Proof. Starting from the known equation
In circular motion, we have v = ω × R. Thus,
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.
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
where M1 is the mass of planet and M2 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:
Therefore, using the Third Kepler Law (the Law of Periods), we obtain for mass of the Moon M2,
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:
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
As you can see, the unit of gravitational potential is [m2 / s2], 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
where m is mass of the object at the position R inside the gravitational field which is affected by the gravitational potential ϕ.
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