Science Part i Solutions Solutions for Class 10 Science Chapter 10 - Space Missions

Answer:

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.

Answer:

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.

Answer:

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:

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. 

Answer:

Answer:

a. Escape velocity for Earth is
$v_{\mathrm{esc}}=\sqrt{\frac{2G{M}_{\mathrm{e}}}{R_{\mathrm{e}}}}=11.2 \mathrm{km}/\mathrm{s}$
Given:
Mass of the planet, M_p = 8$\times$Mass of the Earth (M_e)
Radius of the planet, R_p = 2$\times$Radius of the Earth (R_e)
Thus, escape velocity of the planet is
$v_{\mathrm{esc}}^{\text{'}}=\sqrt{\frac{2G{M}_{\mathrm{p}}}{R_{\mathrm{p}}}}=\sqrt{\frac{8}{2}}\sqrt{\frac{2G{M}_{\mathrm{e}}}{R_{\mathrm{e}}}}=\sqrt{4}\times 11.2=22.4 \mathrm{km}/\mathrm{s}$

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, v_c is given as
$v_{\mathrm{c}}=\sqrt{\frac{GM}{R+h}}$
Thus, 
$$\begin{gathered} T=\frac{2\pi (R+h)}{\sqrt{\frac{GM}{R+h}}}=\frac{2\pi (R+h)\sqrt{R+h}}{\sqrt{GM}} .....\left(\mathrm{i}\right) \\ \Rightarrow T\propto \frac{1}{\sqrt{M}} .....\left(\mathrm{ii}\right) \end{gathered}$$
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 = $6\times {10}^{24} \mathrm{kg}$
R = $6.4\times {10}^5 \mathrm{m}$
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
$T_{4M}=\frac{24}{2}=12 \mathrm{h}$

c. Time period of the satellite is given as
$T=\frac{2\pi (R+h)\sqrt{R+h}}{\sqrt{GM}}$
When the height of the satellite is h₁, it takes T time to revolve around the Earth. Thus,
$T=\frac{2\pi (R+h_1)\sqrt{R+h_1}}{\sqrt{GM}}=\frac{2\pi (R+h_1{)}^{{\displaystyle \frac{3}{2}}}}{\sqrt{GM}}$   .....(i)
When the satellite takes $2\sqrt{2} \mathrm{T}$ time to revolve around the Earth, let it be at height h₂. Thus,
 $2\sqrt{2}T=\frac{2\pi (R+h_2)\sqrt{R+h_2}}{\sqrt{GM}}=\frac{2\pi {\left(R+h_2\right)}^{\frac{3}{2}}}{\sqrt{GM}}$   .....(ii)
Dividing (ii) by (i), we get
$$\begin{gathered} 2\sqrt{2}=\frac{{\left(R+h_2\right)}^{{\displaystyle \frac{3}{2}}}}{{\left(R+h_1\right)}^{\frac{3}{2}}} \\ \Rightarrow \frac{R+h_2}{R+h_1}=2 \\ \Rightarrow h_2=R+2h_1 \end{gathered}$$