Goal: learn to calculate the period of revolution of a satellite around a planet depending on its mass, size and type of satellite.
Work progress:
1. Draw the table presented at the bottom of the table into your notebook.
2. Calculate the orbital period for each satellite for each planet and present the result in the table on the page. It is known that a planet that is 2 times heavier than the Earth is 1.4 times larger in size, and a planet that is smaller than the Earth in mass is 0.8 times the size of the Earth. Data must be taken from the information window on the “Simulation of satellite motion” page. The radius of the Earth is taken to be 6400 km. The answer should be expressed in minutes, rounded to the nearest whole number.
3. Check the data you received. To do this, click the "Check Results" button.
4. If there are errors, correct them.
5. Write down the correct data obtained in a table in your notebook.
6. Draw a conclusion about how the satellite’s orbital period depends on the size of the planet and the type of satellite.
2.2.2. Movement under influence gravity (satellites)
When satellites move (with the engine turned off) in a circular orbit, only one force acts on them - the force of attraction of the satellite to the planet.
A satellite having a mass m and moving in a circular orbit at a height h above the surface of the planet (Fig. 2.2) is affected only by the force of gravity.
Rice. 2.2
This force is directed towards the center of the planet and imparts centripetal acceleration to the satellite. In this case the relation is valid
G m M r 2 = m v 2 r,
allowing us to obtain a formula for calculation escape velocity satellite:
where G = 6.67 ⋅ 10 −11 N ⋅ m 2 /kg 2 - universal gravitational constant; m - body weight; r = R + h - orbital radius; R is the radius of the planet; h is the height of the satellite above the surface of the planet.
There are first, second and third cosmic velocities. For planet Earth:
- first escape velocity- the minimum speed imparted to the satellite near the Earth's surface at which it can enter a circular orbit and begin rotating around the Earth in low-Earth orbit (h ≈ 0),
v 1 ≈ 7.9 km/s;
- second escape velocity- the minimum speed imparted to a satellite near the surface of the Earth at which it can move away from the Earth to a great distance and become a satellite of the Sun,
v 2 ≈ 11.2 km/s;
- third escape velocity- the minimum speed reported to the satellite near the Earth's surface at which it can leave the Solar System; its value is approximately 16.6 km/s.
When they talk about the first escape velocity for a planet, they mean that the satellite is moving at an altitude h ≈ 0, i.e. The radius of the satellite's orbit r coincides with the radius of the planet R:
r = R.
Satellite orbital period around the planet (time of one revolution) can be defined as the ratio of the orbital length to the first escape velocity:
where L = 2πr is the length of the orbit with radius r (circumference); v is the first escape velocity of the satellite in this orbit.
Example 5. How many times does the orbital period of an artificial satellite moving in a circular orbit at an altitude equal to twice the radius of the Earth exceed the orbital period of a satellite rotating in near-Earth orbit?
Solution. The orbital period of a satellite moving in a circular orbit at an altitude h 1 = 2R is determined by the formula
T 1 = 2 π (R + h 1) v 1,
where R is the radius of the Earth; v 1 is the first escape velocity of the satellite at altitude h 1 .
The orbital period of a satellite moving in low-Earth orbit (h 2 ≈ 0) is determined by the formula
T 2 = 2 π (R + h 2) v 2,
where v 2 is the first escape velocity of the satellite in low-Earth orbit.
Substituting the values h 1 = 2R and h 2 = 0 into the formula for calculating the corresponding periods gives:
T 1 = 6 π R v 1 and T 2 = 2 π R v 2 .
Period ratio
T 1 T 2 = 3 v 2 v 1
is expressed through the ratio of the first cosmic velocities of the satellite in the corresponding orbits.
The first cosmic velocities are determined by the following formulas:
- for height h 1 = 2R
v 1 = G M R + h 1 = G M R + 2 R = G M 3 R ;
- for height h 2 ≈ 0 (earth orbit)
v 2 = G M R + h 2 = G M R + 0 = G M R ,
where G = 6.67 ⋅ 10 −11 N m 2 /kg 2 - universal gravitational constant; M is the mass of the Earth.
Substituting v 1 and v 2 into the formula for the ratio of periods, we get
T 1 T 2 = 3 v 2 v 1 = 3 G M R ⋅ 3 R G M = 3 3 ≈ 5.2.
those. The orbital period of a satellite moving at an altitude equal to two radii exceeds the orbital period of a satellite in low-Earth orbit by approximately 5.2 times.
Example 6. The radius of a certain planet is 3 times greater than the radius of the Earth, and its density is 9 times less than the density of the Earth. Determine the ratio of the first cosmic velocities of satellites for the Earth and for the planet.
Solution. The following first escape velocities are compared:
- for the Earth's surface
v 1 = G M Z R Z,
- for the surface of the planet
where G = 6.67 ⋅ 10 −11 N m 2 /kg 2 - universal gravitational constant; MZ - mass of the Earth; RZ - radius of the Earth; M is the mass of the planet; R is the radius of the planet.
The speed ratio is
v 1 v 2 = M Z R Z R M .
Assuming that the Earth and the planet have a spherical shape, we obtain formulas for calculating the corresponding masses:
- for the Earth
M Z = ρ Z V Z = 4 3 π ρ Z R Z 3,
- for the planet
M = ρ V = 4 3 π ρ R 3 ,
where ρ Z is the density of the Earth; ρ is the density of the planet.
Let's substitute the expressions for the masses into the formula for the speed ratio:
v 1 v 2 = 4 3 π ρ З R З 3 R З 3 4 R π ρ R 3 = ρ З R З 2 ρ R 2 = R З R ρ З ρ .
According to the conditions of the problem, R = 3R З and ρ З = 9ρ; therefore, the required speed ratio is equal to
v 1 v 2 = R З 3 R З 9 ρ ρ = 1,
those. the satellite speeds are the same for the surface of the Earth and for the surface of the planet.
Example 7. A satellite rotates around a certain planet in a circular orbit with a radius of 20,000 km at a speed of 12 km/s. Determine the magnitude of the acceleration due to gravity on the surface of the planet if its radius is 12,000 km.
Solution. We find the acceleration of free fall on the surface of the planet using the formula
where G = 6.67 ⋅ 10 −11 N m 2 /kg 2 - universal gravitational constant; M is the mass of the planet; R is the radius of the planet.
The radius of the planet is specified in the problem statement; the product (GM) can be expressed from the formula for the first escape velocity:
v = G M R + h = G M r ,
where r is the radius of the satellite’s orbit; hence the required work
GM = v 2 r.
Let's substitute (GM) into the expression to calculate g 0:
g 0 = v 2 r R 2 .
The calculation allows us to obtain the value of the acceleration of gravity on the surface of the planet:
g 0 = (12 ⋅ 10 3) 2 ⋅ 2, 0 ⋅ 10 7 (12 ⋅ 10 6) 2 = 20 m/s 2.
In space, gravity provides the force that causes satellites (such as the Moon) to orbit larger bodies (such as the Earth). These orbits generally have the shape of an ellipse, but most often this ellipse is not very different from a circle. Therefore, to a first approximation, the orbits of satellites can be considered circular. Knowing the mass of the planet and the height of the satellite’s orbit above the Earth, we can calculate what it should be speed of the satellite around the Earth.
Calculation of the speed of a satellite around the Earth
Rotating in a circular orbit around the Earth, a satellite at any point in its trajectory can only move at a constant absolute speed, although the direction of this speed will constantly change. What is the magnitude of this speed? It can be calculated using Newton's second law and the law of gravity.
To maintain a circular orbit of a mass satellite in accordance with Newton's second law, a centripetal force will be required: , where is the centripetal acceleration.
As is known, centripetal acceleration is determined by the formula:
where is the speed of the satellite, is the radius of the circular orbit along which the satellite moves.
Centripetal force is provided by gravity, therefore, in accordance with the law of gravity:
where kg is the mass of the Earth, m 3 ⋅kg -1 ⋅s -2 is the gravitational constant.
Substituting everything into the original formula, we get:
Expressing the required speed, we find that the speed of the satellite around the Earth is equal to:
This is a formula for the speed that an Earth satellite must have at a given radius (i.e. distance from the center of the planet) to maintain a circular orbit. The speed cannot change in magnitude as long as the satellite maintains a constant orbital radius, that is, as long as it continues to orbit the planet in a circular path.
When using the resulting formula, there are several details to consider:
Artificial satellites of the Earth, as a rule, orbit the planet at an altitude of 500 to 2000 km from the surface of the planet. Let's calculate the speed at which such a satellite should move at an altitude of 1000 km above the Earth's surface. In this case km. Substituting the numbers, we get:
Material prepared by Sergei Valerievich
How many times does the period of revolution of an artificial satellite moving in a circular orbit at an altitude equal to the radius of the Earth exceed the period of revolution of a satellite in low-Earth orbit?
Problem No. 2.5.14 from the “Collection of problems for preparing for entrance exams in physics at USPTU”
Given:
\(h=R\), \(\frac(T_2)(T_1)-?\)
Solution to the problem:
Let us find the orbital period \(T_2\) of a satellite moving in a circular orbit at an altitude \(h=R\). It is clear that the force of universal gravity imparts to the satellite centripetal acceleration \(a_t\), therefore Newton’s second law will be written in the following form:
\[(F_(t2)) = m(a_(t2))\;\;\;\;(1)\]
The force of gravity is determined by the law of universal gravitation:
\[(F_(t2)) = G\frac((Mm))((((\left((R + h) \right))^2)))\;\;\;\;(2)\ ]
In order for the period of revolution to appear in our formula, we need to express the centripetal acceleration \(a_(c2)\) through it. To do this, we write the formula for determining the acceleration \(a_(q2)\) through the angular velocity and the formula for connecting the latter with the period.
\[(a_(t2)) = (\omega ^2)\left((R + h) \right)\]
\[\omega = \frac((2\pi ))(T_2)\]
\[(a_(t2)) = \frac((4(\pi ^2)))(T_2^2)\left((R + h) \right)\;\;\;\;(3)\ ]
Let's substitute expressions (2) and (3) into equality (1):
Let's make an analogy for a satellite moving in low-Earth orbit. It is clear that its revolution period will be equal to:
\[(T_1) = 2\pi \sqrt (\frac(((R^3)))((GM)))\]
Now let’s substitute the condition \(h=R\) into the formula for determining the period \(T_2\) (in formula (4)):
\[(T_2) = 2\pi \sqrt (\frac((((\left((R + R) \right))^3)))((GM))) = 2\pi \sqrt (\frac ((8(R^3)))((GM))) \]
The required ratio is:
\[\frac(((T_2)))(((T_1))) = \sqrt 8 = 2\sqrt 2 = 2.83\]
Answer: 2.83 times.
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