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Question 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 ...... .
b. The initial velocity (during launching) of the Managalyaan, must be greater than ..............of the earth.

a. If the height of the orbit of a satellite from the earth surface is increased, the tangential velocity of the satellite will decrease.
Explanation: This can be easily inferred from the formula of critical or tangential velocity.
${v}_{\mathrm{c}}=\sqrt{\frac{GM}{R+h}}$

b. The initial velocity (during launching) of the Managalyaan, must be greater than escape velocity of the earth.

Question 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.
b. The escape velocity on the moon is less than that on the earth.
c. A satellite needs a specific velocity to revolve in a specific orbit.
d. If the height of the orbit of a satellite increases, its velocity must also increase.

a. The statement is false. This is because the minimum velocity with which the spacecraft must be projected so that it escapes the Earth's gravitational pull is known as escape velocity. So, the initial velocity of the spacecraft must be greater than or equal to escape velocity of Earth.

b. Escape velocity on a planet is given as:
${v}_{\mathrm{esc}}=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$
Since, acceleration due to gravity on Moon is one-sixth of that on Earth, thus the escape velocity on the Moon is less than that on the Earth. Hence, the statement is true.

c. The specific velocity with which the satellite revolves around a planet is known as critical velocity. It is given as
${v}_{\mathrm{c}}=\sqrt{\frac{GM}{R+h}}$
So, it can be observed that the critical velocity of a satellite changes depending on the height of the orbit from the surface of a planet. Thus, to revolve in a specific orbit, a satellite would require specific velocity. Hence, the statement is true.

d. The specific velocity with which the satellite revolves around a planet is known as critical velocity. It is given as
${v}_{\mathrm{c}}=\sqrt{\frac{GM}{R+h}}$
So, we see that as the height (h) of the orbit of a satellite increases, it velocity must decrease. Hence, the statement is false.

Question 3:

Answer the following questions:
a. What is meant by an artificial satellite? How are the satellites classified based on their functions?
b. What is meant by the orbit of a satellite? On what basis and how are the orbits of artificial satellites classified?
c. Why are geostationary satellites not useful for studies of polar regions?
d. What is meant by satellite launch vehicles? Explain a satellite launch vehicle developed by ISRO with the help of a schematic diagram.
e. Why it is beneficial to use satellite launch vehicles made of more than one stage?

a. A man made object revolving around a planet in a fixed orbit is known as artificial satellites. Based on their functions, satellites are classified as following:

 Type of satellite Function of satellite Weather satellite Study and prediction of weather Communication satellite Establish communication between different location in the world through use of specific waves Broadcast satellite Telecasting of television programs Navigational satellite Fix the location of any place on the Earth’s surface in terms of its very precise latitude and longitude Military Satellite Collect information for security aspects Earth Observation Satellite Study of forests, deserts, oceans, polar ice on the earth’s surface, exploration and management of natural resources, observation and guidance in case of natural calamities like flood and earthquake

b. An orbit is a specific path (elliptical or circular) or trajectory around a planet in which a satellite revolves. â€‹Depending on the height of the satellite’s orbit above the Earth’s surface, the satellite orbits are classified as below:
• High Earth Orbits (Height from the earth’s surface > 35780 km): If the height of the satellite’s orbit above the earth’s surface is greater than or equal to 35780 km, the orbit is called High earth Orbit
• Medium Earth Orbit (Height above the earth’s surface 2000 km to 35780 km): If the height of the satellite orbit above the earth’s surface is in between 2000 km and 35780 km, the orbits are called medium earth orbits.
• Low Earth Orbits (Height above the earth’s surface: 180 km to 2000 km): If the height of the satellite orbit above the earth’s surface is in between 180 km and 2000 km, the orbits are called Low earth Orbits.

c. The geostationary satellites orbit above the equator. Thus, these are not useful for studies of polar regions.

d. Satellite launch vehicles are used to place the satellites in their specific orbits. The structure of polar satellite launch vehicle (PSLV) developed by ISRO is shown below. It is a vehicle with 4 stages using solid and liquid fuels. The weight of the vehicle decreases after each stage because of consumption of fuel at that stage and detachment of that stage (i.e. the empty tank) from its body. Thus, the vehicle moves with higher speed after every stage. e. The major portion of weight in satellite launch vehicles is contributed by fuel. Thus, vehicles have to carry a large weight of fuel during their course of journey. To overcome this problem of launch vehicles carrying heavy load during its entire journey, it is provided with more than one stage. Because to this, the weight of the vehicle can be reduced step by step, after its launching. 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. 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 a sea or on an unpopulated land. As the fuel in the first stage is exhausted, the fuel in the second stage is ignited. However, the vehicle now contains only one (i.e. the second) stage. The vehicle can move with higher speed as the weight has reduced.

Question 4:

Complete the following table.  Question 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?
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?
c. If the height of a satellite completing one revolution around the earth in T seconds is h1 meter, then what would be the height of a satellite taking seconds for one revolution?

a. Escape velocity for Earth is

Given:
Mass of the planet, Mp = 8$×$Mass of the Earth (Me)
Radius of the planet, Rp = 2$×$Radius of the Earth (Re)
Thus, escape velocity of the planet is

b. Given:
Height of the satellite, h = 35780 km
Let the original mass of Earth be M. Then its new mass will be 4M.
Time taken by the satellite to revolved around the Earth's orbit is given as
$T=\frac{2\mathrm{\pi R}}{{v}_{c}}$
Now, vc is given as
${v}_{\mathrm{c}}=\sqrt{\frac{GM}{R+h}}$
Thus,

Thus from equation (ii) we see that when the mass of the Earth becomes 4 times, the time period of revolution of satellite should be halved.
i.e. ${T}_{4M}=\frac{T}{2}$     .....(iii)
Now, h = 35780 km
M
R
G = 6.67 × 10-11 N m2 /kg2
Putting the values of h, M and R in first, we get
T = 24 h
Using (iii), we get

c. Time period of the satellite is given as
$T=\frac{2\pi \left(R+h\right)\sqrt{R+h}}{\sqrt{GM}}$
When the height of the satellite is h1, it takes T time to revolve around the Earth. Thus,
$T=\frac{2\pi \left(R+{h}_{1}\right)\sqrt{R+{h}_{1}}}{\sqrt{GM}}=\frac{2\pi \left(R+{h}_{1}{\right)}^{\frac{3}{2}}}{\sqrt{GM}}$   .....(i)
When the satellite takes  time to revolve around the Earth, let it be at height h2. Thus,
$2\sqrt{2}T=\frac{2\pi \left(R+{h}_{2}\right)\sqrt{R+{h}_{2}}}{\sqrt{GM}}=\frac{2\pi {\left(R+{h}_{2}\right)}^{\frac{3}{2}}}{\sqrt{GM}}$   .....(ii)
Dividing (ii) by (i), we get
$2\sqrt{2}=\frac{{\left(R+{h}_{2}\right)}^{\frac{3}{2}}}{{\left(R+{h}_{1}\right)}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}⇒\frac{R+{h}_{2}}{R+{h}_{1}}=2\phantom{\rule{0ex}{0ex}}⇒{h}_{2}=R+2{h}_{1}$

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