Chapter 10 – Space Missions
1. Fill in the blanks and explain the statements with reasoning:
a. If the height of the orbit of a satellite from the earth surface is increased, the tangential velocity of the satellite will ______.
Ans: decrease
b. The initial velocity (during launching) of the Mangalyaan, must be greater than ______ of the earth.
Ans: escape velocity
2. State with reasons whether the following sentences are true or false
a. If a spacecraft has to be sent away from the influence of earth’s gravitational field, its velocity must be less than the escape velocity.
Ans: False.
Explanation: The escape velocity is responsible for launching a rocket in space which can overcome the earth’s gravitational force. Hence, its velocity should be more than the escape velocity.
b. The escape velocity on the moon is less than that on the earth.
Ans: True.
Explanation: As the gravitational force of attraction on the moon is 1/6th of the Earth, hence, less escape velocity is required to overcome the gravitational force.
c. A satellite needs a specific velocity to revolve in a specific orbit.
Ans: True
d. If the height of the orbit of a satellite increases, its velocity must also increase.
Ans: False.
Explanation: If height increases, the velocity decreases.
3. Answer the following questions:
a. What is meant by an artificial satellite? How are the satellites classified based on their functions?
Ans: A manmade object revolves around the earth or any other planet in a fixed orbit is called an artificial satellite. Satellites work on solar energy and hence photovoltaic panels are attached on both sides of the satellite, which look like wings. Satellites are also installed with various transmitters and other equipment to receive and transmit signals between the earth and the satellites.
Classification of satellites depending on their functions:
a. Weather satellite:
Weather satellites collect information regarding the weather conditions in the region.
b. Communication satellite:
In order to establish communication between different places on the earth through mobile phones or computer-assisted internet, communication satellites are used.
c. Broadcast satellite:
Broadcast satellites are used to transmit various radio and television programs and even live programs from any place on earth to any other place.
d. Navigational satellite:
Navigational satellites assist surface, water, and air transportation and coordinate their busy schedules.
e. Military satellite:
Military satellites help monitor all movements of neighbouring or enemy countries by providing real-time information about the borders. They also help guide the missiles effectively.
f. Earth observation satellites:
These satellites observe and provide real-time information about the earth.
b. What is meant by the orbit of a satellite? On what basis and how are the orbits of artificial satellites classified?
Ans:
(i) The particular path that a satellite follows to revolve around the earth is known as its orbit. Different artificial satellites revolve in different orbits around the earth.
(ii) The orbits of artificial satellites are classified on the basis of following factors:
(a) Height of the orbit above the Earth’s surface.
(b) The shape of the orbit (i.e. circular or elliptical)
(c) Whether it is parallel to the equator or makes a particular angle with the equator.
(iii) According to the altitude, the orbits are classified as follows:
(a) High Earth Orbits: Height 35780 km or higher above the Earth’s surface.
(b) Medium Earth Orbits: Height 2000 km to 35780 km above the Earth’s surface.
(c) Low Earth Orbits: Height 180 km to 2000 km, above the Earth’s surface.
c. Why are geostationary satellites not useful for studies of polar regions?
Ans:
(i) Geo-stationary satellites have orbits parallel to the equator and hence the observation of polar regions is not carried out properly.
(ii) For this purpose, elliptical medium earth orbits passing over polar regions are used.
d. What is meant by satellite launch vehicles? Explain a satellite launch vehicle developed by ISRO with the help of a schematic diagram.
Ans:
(i) Satellite launch vehicles are used to place the satellites in their specific orbits. The functioning of the satellite launch vehicle is based on Newton’s third law of motion.
(ii) The launch vehicle uses a specific type of fuel.
(iii) The gas produced due to combustion of the fuel expands due to its high temperature and is expelled forcefully through the nozzles at the rear of the launch vehicle.
(iv) As a reaction to this, a thrust acts on the vehicle, which drives the vehicle high into space.
(v) The structure of the launch vehicle is decided by the weight of the satellite and the type of satellite orbit.
(vi) The fuel forms a major portion of the total weight of the launch vehicle.
(vii) To overcome this problem, launch vehicles with more than one stage are used. Due to this, the weight of the vehicle can be reduced step by step, after its launching.
(viii) For example, consider a launch vehicle having two stages. For launching the vehicle, the fuel and engine in the first stage are used. This imparts a specific velocity to the vehicle and takes it to a certain height.
(ix) Once the fuel in this first stage is exhausted, the empty fuel tank and the engine are detached from the main body of the vehicle and fall either into the sea or on unpopulated land.
(x) As the fuel in the first stage is exhausted, the fuel in the second stage is ignited.
(xi) Since, the vehicle now contains only one stage, the weight has now been reduced and the vehicle can move with higher speed. Almost all vehicles are made of either two or more stages.
(xii) As an example, Polar Satellite Launch Vehicle (PSLV).
e. Why it is beneficial to use satellite launch vehicles made of more than one stage?
Ans:
(i) Satellite Launch Vehicle uses a special type of fuel in huge amounts.
(ii) Hence, the maximum weight of the launcher is due to the fuel.
(iii) Due to this, different stages of fuels are made so that as the fuel gets utilized, the weight of the satellite launch vehicle can gradually reduce.
(iv) As the fuel in the 1st step is burnt, that part gets automatically detached from the launcher. Due to this, the satellite launch vehicle can acquire more velocity.
(v) Thus, ultimately escape velocity is achieved and satellite can be launched successfully.
4. Complete the following table.
Ans:
5. Solve the following problems.
a. If mass of a planet is eight times the mass of the earth and its radius is twice the radius of the earth, what will be the escape velocity for that planet?
Given:
Mass on planet (\(M_{p}\)) = 8 \(M_{e}\)
Radius of planet (\(R_{p}\)) = 2 \(R_{e}\)
To find:
Escape velocity for the planet (\(V_{e(p)}\))
Solution:
(a) For Earth,
\(V_{e}\) = \(\sqrt {\large \frac {2GM_{e}}{R_{e}}}\) = 11.2 km/s …(i)
(b) For the given planet,
\(V_{c}\) = \(\sqrt {\large \frac {2GM_{p}}{R_{p}}}\) …(ii)
∴ \(V_{c}\) = \(\sqrt {\large \frac {2G\,×\,8\,×\,M_{e}}{2\,×\,R_{e}}}\) …[∵ (\(M_{p}\)) = 8 (\(M_{e}\)) & (\(R_{p}\)) = 2 (\(R_{e}\))]
∴ \(V_{c}\) = \(\sqrt {\large \frac {2G\,×\,4\,×\,M_{e}}{R_{e}}}\)
∴ \(V_{c}\) = 2\(\sqrt {\large \frac {2GM_{e}}{R_{e}}}\)
∴ \(V_{c}\) = 2 × 11.2 …[from (i)]
∴ \(V_{c}\) = 22.4 km/s
Ans: The escape velocity of the planet is 22.4 km/s.
b. How much time a satellite in an orbit at height 35780 km above earth’s surface would take, if the mass of the earth would have been four times its original mass?
Given:
h = 35780 km
v = 3.08 km/s
\(M_{n}\) = 4 M [∵ \(M_{n}\) is the new Mass]
R = 6400 km
To find:
\(T_{n}\) = ?
Solution:
We know that,
v = \(\sqrt {\large \frac {GM}{R\,+\,h}}\) …(i)
Let \(v_{n}\) be the new velocity
∴ \(v_{n}\) = \(\sqrt {\large \frac {GM_{n}}{R\,+\,h}}\)
∴ \(v_{n}\) = \(\sqrt {\large \frac {G\,×\,4M}{R\,+\,h}}\)
∴ \(v_{n}\) = 2 × \(\sqrt {\large \frac {GM}{R\,+\,h}}\)
∴ \(v_{n}\) = 2 × v …(ii) [from (i)]
Also,
\(v_{n}\) = 2π × \(\large \frac {R\,+\,h}{T_{n}}\)
∴ \(T_{n}\) = 2π × \(\large \frac {R\,+\,h}{v_{n}}\)
∴ \(T_{n}\) = 2π × \(\large \frac {R\,+\,h}{2v}\) …[from (i)]
∴ \(T_{n}\) = π × \(\large \frac {R\,+\,h}{v}\)
∴ \(T_{n}\) = 3.14 × \(\large \frac {35780\,+\,6400}{3.08}\)
∴ \(T_{n}\) = 3.14 × \(\large \frac {42180}{3.08}\)
∴ \(T_{n}\) = \(\large \frac {132455.2}{3.08}\)
∴ \(T_{n}\) = 43001.68 s
We have to convert it in hours,
∴ \(T_{n}\) = \(\large \frac {43001.68}{3600}\)
∴ \(T_{n}\) = 11.94 hours ~ 12 hours
Ans: The time taken will be 12 hours.
c. If the height of a satellite completing one revolution around the earth in T seconds is h\(_{1}\) meter, then what would be the height of a satellite taking 2 \(\sqrt {T}\) seconds for one revolution?
Given:
Height of 1st satellite = h\(_{1}\)
Time of rev. of 1st satellite = T
Time of rev. of 2nd satellite = 2 \(\sqrt {2T}\)
To find:
Height of 2nd satellite (h\(_{2}\)) = ?
Solution:
We know that,
\(\large \frac {T²}{R³}\) = k
Case (i)
Time = T
r = R + h\(_{1}\)
∴ \(\large \frac {T²}{(R\,+\,h_{1})³}\) = k
Case (ii)
Time = 2 \(\sqrt {2T}\)
r = R + h\(_{2}\)
∴ \(\large \frac {8T²}{R\,+\,h_{2}}\)³ = k
From (i) & (ii),
\(\large \frac {T²}{(R\,+\,h_{1})³}\) = \(\large \frac {8T²}{R\,+\,h_{2}}\)³
∴ \((R\,+\,h_{2})\)³ = 8 \((R\,+\,h_{1})\)³
Taking cube roots on both sides, we get,
\(\sqrt[3] {(R\,+\,h_{2})³}\) = \(\sqrt[3] {8(R\,+\,h_{1})³}\)
∴ R + h\(_{2}\) = 2 (R + h\(_{1}\))
∴ R + h\(_{2}\) = 2R + 2h\(_{1}\)
∴ h\(_{2}\) = 2R + 2h\(_{1}\) – R
∴ h\(_{2}\) = R + 2h\(_{1}\)
Ans: Height of the second satellite is R + 2h\(_{1}\).