Space Missions Class 10th Solutions | Space Missions SSC Class 10 Questions And Answers
Fill in the blank and explain the statement with reasoning:
If the height of the orbit of a satellite from the Earth's surface is increased, the tangential velocity of the satellite will ______.
Solution: decrease.
Fill in the blank and explain the statement with reasoning:
The initial velocity (during launching) of the Mangalyaan must be greater than ______ of the earth.
Solution 1: Scientific and Written Exam Answer:
The initial velocity of Mangalyaan must be greater than the escape velocity of Earth. The escape velocity is given by:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
For Earth, the escape velocity is approximately 11.2 km/s. To reach Mars, Mangalyaan needed a velocity greater than this to overcome Earth's gravitational pull and enter a transfer orbit.
Solution 2: Simple and Understandable Answer:
To escape Earth's gravity and travel toward Mars, Mangalyaan needed a speed greater than 11.2 km/s. This is called escape velocity, which allows an object to move away from Earth without falling back due to gravity.
State with reasons whether the sentence is true or false:
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.
Solution 1: Scientific and Written Exam Answer:
False. The escape velocity is the minimum velocity required for an object to completely escape Earth's gravitational influence. It is given by:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
For Earth, the escape velocity is 11.2 km/s. If the spacecraft’s velocity is less than this value, Earth's gravity will pull it back, and it will not be able to escape into space.
Solution 2: Simple and Understandable Answer:
False. To leave Earth's gravity and travel into space, a spacecraft needs a speed of at least 11.2 km/s. If it moves slower than this, Earth's gravity will pull it back, preventing it from escaping.
---State with reasons whether the sentence is true or false:
The escape velocity on the moon is less than that on the earth.
Solution 1: Scientific and Written Exam Answer:
True. The escape velocity is directly related to the mass and radius of a celestial body. It is given by:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
The Moon has a much smaller mass and a lower gravitational pull than Earth. As a result, its escape velocity is only 2.38 km/s, which is significantly lower than Earth's escape velocity of 11.2 km/s.
Solution 2: Simple and Understandable Answer:
True. The Moon has weaker gravity than Earth because it is much smaller in size. This means that objects on the Moon need much less speed (2.38 km/s) to escape compared to Earth (11.2 km/s).
State with reasons whether the sentence is true or false:
A satellite needs a specific velocity to revolve in a specific orbit.
Solution 1: Scientific and Written Exam Answer:
True. A satellite revolves around the Earth due to the balance between gravitational force and the centrifugal force. The velocity required for a satellite to stay in a stable orbit is called orbital velocity and is given by:
$$ v_o = \sqrt{\frac{GM}{R+h}} $$
where,
- G = Gravitational constant
- M = Mass of the Earth
- R = Radius of the Earth
- h = Height of the satellite from the Earth’s surface
If the velocity is too low, the satellite will fall back to Earth. If it is too high, it will escape into space.
Solution 2: Simple and Understandable Answer:
True. A satellite must travel at the correct speed to stay in orbit. If it goes too slow, gravity will pull it back to Earth. If it goes too fast, it will escape into space.
State with reasons whether the sentence is true or false:
If the height of the orbit of a satellite increases, its velocity must also increase.
Solution 1: Scientific and Written Exam Answer:
False. The orbital velocity of a satellite is given by:
$$ v_o = \sqrt{\frac{GM}{R+h}} $$
From this equation, we can see that as the height (h) of the satellite increases, the denominator increases, reducing the velocity (v_o). Hence, a satellite in a higher orbit moves slower than one in a lower orbit.
Solution 2: Simple and Understandable Answer:
False. If a satellite is placed in a higher orbit, it moves slower, not faster. This is because gravity is weaker at higher altitudes, so the satellite needs less speed to stay in orbit.
Answer the question
What is meant by an artificial satellite? How are the satellites classified based on their functions?
Solution 1: Scientific and Written Exam Answer
An artificial satellite is a human-made object that is placed into orbit around a celestial body, typically the Earth, for various scientific, communication, or military purposes. These satellites are launched using rockets and follow specific orbital paths.
Classification of Satellites Based on Their Functions:
- Communication Satellites: Used for transmitting television, radio, and internet signals (e.g., INSAT series).
- Weather Satellites: Used for monitoring weather patterns, climate changes, and natural disasters (e.g., METSAT).
- Navigation Satellites: Provide GPS services for location tracking and navigation (e.g., NAVIC, GPS).
- Scientific Satellites: Used for space research and astronomical observations (e.g., Hubble Space Telescope).
- Military Satellites: Used for defense, surveillance, and intelligence gathering (e.g., Spy Satellites).
- Remote Sensing Satellites: Used for observing Earth's surface for agricultural, geographical, and environmental studies (e.g., Cartosat).
Solution 2: Simple and Understandable Answer
An artificial satellite is a machine made by humans that is sent into space to orbit the Earth or another planet. It helps in communication, weather forecasting, navigation, and research.
Types of Satellites:
- Communication Satellites: Help in TV, radio, and internet signals.
- Weather Satellites: Track storms, rains, and climate changes.
- Navigation Satellites: Help in GPS and maps.
- Scientific Satellites: Study space and planets.
- Military Satellites: Used for defense and spying.
- Remote Sensing Satellites: Monitor Earth’s land, water, and forests.
For example, the INSAT satellite helps in communication and weather forecasting in India.
What is meant by the orbit of a satellite? On what basis and how are the orbits of artificial satellites classified?
Solution 1: Scientific and Written Exam Answer
The orbit of a satellite is the curved path in which the satellite moves around a celestial body, such as the Earth, due to the gravitational force acting on it. The orbit depends on the satellite’s velocity, altitude, and the gravitational pull of the central body.
Classification of Satellite Orbits:
Orbits of artificial satellites are classified based on altitude, inclination, and function.
- Based on Altitude:
- Low Earth Orbit (LEO): Altitude between 180 km - 2000 km (e.g., International Space Station).
- Medium Earth Orbit (MEO): Altitude between 2000 km - 35,786 km (e.g., GPS satellites).
- Geostationary Orbit (GEO): Altitude of 35,786 km, remains fixed relative to Earth (e.g., Communication satellites).
- Based on Inclination:
- Polar Orbit: Passes over the poles, covering the entire Earth (e.g., Remote sensing satellites).
- Equatorial Orbit: Orbits along the equator.
- Based on Function:
- Sun-Synchronous Orbit: Passes over the same area at the same local solar time (e.g., Weather satellites).
- Transfer Orbits: Used to move satellites from one orbit to another.
Solution 2: Simple and Understandable Answer
The orbit of a satellite is the path it follows around a planet due to gravity. It is like a road in space that a satellite travels on.
Types of Orbits:
- Low Earth Orbit (LEO): Close to Earth, used for space stations and weather satellites.
- Medium Earth Orbit (MEO): Higher up, mainly used for GPS.
- Geostationary Orbit (GEO): Very high, used for TV and communication satellites.
- Polar Orbit: Moves from pole to pole, covering the entire planet.
- Sun-Synchronous Orbit: Passes over the same place at the same time each day, useful for weather tracking.
For example, the GPS satellites use MEO to help with navigation.
Why are geostationary satellites not useful for studies of polar regions?
Solution 1: Scientific and Written Exam Answer
Geostationary satellites are positioned in a geostationary orbit at an altitude of approximately 35,786 km above the Earth's equator. These satellites rotate with the Earth and remain fixed over a specific location.
Reason:
- Geostationary satellites orbit along the equatorial plane and do not move over the poles.
- Since their line of sight remains fixed near the equator, they have poor visibility of high-latitude and polar regions.
- As a result, they cannot effectively monitor the weather, climate, or environmental changes in the polar areas.
To study polar regions, polar orbiting satellites are used, as they pass over both poles and cover the entire Earth.
Solution 2: Simple and Understandable Answer
Geostationary satellites stay above the equator and do not move towards the poles. Because of this:
- They cannot see the polar regions clearly.
- They are not useful for studying weather, ice melting, or climate changes at the poles.
Instead, polar satellites are used, as they move over the entire Earth, including the poles.
Example: Weather satellites studying the Arctic and Antarctic use polar orbits instead of geostationary orbits.
What is meant by satellite launch vehicles? Explain a satellite launch vehicle developed by ISRO with the help of a schematic diagram.
Solution 1: Scientific and Written Exam Answer
A satellite launch vehicle is a rocket system designed to carry and place satellites into specific orbits. It consists of multiple stages, each providing thrust to push the satellite beyond Earth's atmosphere and into space.
Key Features:
- Uses powerful rocket engines with controlled burning of fuel.
- Has multiple stages to provide enough thrust to reach orbit.
- Designed to deliver satellites into Low Earth Orbit (LEO), Geostationary Orbit (GEO), or other orbits.
Example: PSLV (Polar Satellite Launch Vehicle)
The PSLV, developed by ISRO (Indian Space Research Organisation), is one of the most successful launch vehicles.
PSLV Features:
- Four-stage launch vehicle.
- Capable of placing satellites into polar, geostationary transfer, and low Earth orbits.
- Used for launching remote sensing and communication satellites.
Schematic Diagram of PSLV:
Solution 2: Simple and Understandable Answer
A satellite launch vehicle is a special rocket that carries satellites into space.
How It Works:
- The rocket engine burns fuel to push the satellite upwards.
- It has multiple stages to help the satellite reach space.
- After reaching the required height, it releases the satellite into orbit.
Example: PSLV (Polar Satellite Launch Vehicle)
- It is India’s most reliable rocket, made by ISRO.
- It helps place satellites in different orbits.
- PSLV has been used for famous missions like Chandrayaan and Mangalyaan.
Why is it beneficial to use satellite launch vehicles made of more than one stage?
Solution 1: Scientific and Written Exam Answer
Satellite launch vehicles are designed with multiple stages to efficiently deliver satellites into orbit. Each stage has its own engine and fuel supply, which provides various benefits:
Advantages of Multi-Stage Launch Vehicles:
- Increased Efficiency: Each stage is jettisoned after its fuel is exhausted, reducing the vehicle's weight and allowing the next stage to operate with greater efficiency.
- Higher Velocity: Multi-stage rockets provide the necessary thrust to achieve escape velocity and overcome Earth's gravitational pull.
- Greater Payload Capacity: Since the lower stages drop off, the upper stages can carry heavier payloads into orbit.
- Capability to Reach Different Orbits: Multi-stage rockets can adjust their trajectory for different types of orbits, such as Low Earth Orbit (LEO) and Geostationary Orbit (GEO).
Example: The Polar Satellite Launch Vehicle (PSLV) developed by ISRO has four stages, allowing it to carry multiple payloads efficiently.
Solution 2: Simple and Understandable Answer
Using rockets with more than one stage is beneficial because it makes launching satellites easier and more efficient.
Why is it useful?
- Less Weight: As each stage finishes its fuel, it is dropped off, making the rocket lighter.
- More Power: Different stages give extra thrust to push the satellite higher.
- Carries More Load: With multiple stages, the rocket can carry heavier satellites.
- Reaches Higher Orbits: Multi-stage rockets can send satellites to different altitudes.
Example: India's PSLV rocket has four stages, helping it carry satellites into space effectively.
Solution 1: Scientific and Written Exam Answer
| Satellite Type | Function | Category |
|---|---|---|
| IRNSS | Fixing location of places on Earth's surface using latitude & longitude | Navigational |
| INSAT & GSAT | Weather study & prediction | Weather |
| IRS (Indian Remote Sensing Satellite) | Study of forests, oceans, exploration & management of natural resources | Earth’s observation |
| Communication Satellites | Transmission of TV, radio, and internet signals | Telecommunication |
| Spy Satellites | Military surveillance and security monitoring | Defense & Security |
| Astronomical Satellites | Space observations and study of celestial bodies | Space Research |
| Geostationary Satellites | Remains fixed relative to Earth, used for communication and weather forecasting | Fixed Positioning |
| Polar Satellites | Covers entire Earth, used for environmental and climate monitoring | Earth Monitoring |
| Scientific Satellites | Conducts scientific experiments in space | Research & Development |
Solve the problem.
If the 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?
Solution 1: Scientific and Written Exam Answer
The escape velocity is given by the formula:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
Where:
- G = Gravitational constant
- M = Mass of the planet
- R = Radius of the planet
For Earth, the escape velocity is:
$$ v_{e_{Earth}} = \sqrt{\frac{2G M_{Earth}}{R_{Earth}}} $$
Given:
- Mass of the planet: $$ M_{planet} = 8 M_{Earth} $$
- Radius of the planet: $$ R_{planet} = 2 R_{Earth} $$
Substituting these values:
$$ v_{e_{planet}} = \sqrt{\frac{2G (8M_{Earth})}{2R_{Earth}}} $$
Simplifying:
$$ v_{e_{planet}} = \sqrt{ \frac{8}{2} \times \frac{2G M_{Earth}}{R_{Earth}} } $$
$$ v_{e_{planet}} = \sqrt{4} \times v_{e_{Earth}} $$
$$ v_{e_{planet}} = 2 v_{e_{Earth}} $$
Since the escape velocity of Earth is approximately 11.2 km/s:
$$ v_{e_{planet}} = 2 \times 11.2 $$
$$ v_{e_{planet}} = 22.4 \text{ km/s} $$
Therefore, the escape velocity for the planet is 22.4 km/s.
Solution 2: Simple and Understandable Answer
Escape velocity is the speed needed for an object to leave a planet’s gravity.
The formula for escape velocity is:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
If a planet is 8 times heavier than Earth but has twice its radius, the escape velocity changes.
By substituting the values, we find:
The escape velocity of the planet will be twice that of Earth.
Since Earth's escape velocity is 11.2 km/s, the escape velocity of this planet will be:
22.4 km/s
That means an object must travel at 22.4 km/s to break free from this planet’s gravity!
Solve the problem.
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?
Solution 1: Scientific and Written Exam Answer
The time period of a satellite in orbit is given by Kepler’s third law:
$$ T = 2\pi \sqrt{\frac{r^3}{GM}} $$
Where:
- G = Gravitational constant
- M = Mass of the Earth
- r = Distance from the center of the Earth
The total radius is:
$$ r = R_{Earth} + h $$
Given:
- Height of the satellite: $$ h = 35780 \text{ km} = 3.578 \times 10^7 \text{ m} $$
- Radius of Earth: $$ R_{Earth} = 6370 \text{ km} = 6.37 \times 10^6 \text{ m} $$
So,
$$ r = (6.37 \times 10^6) + (3.578 \times 10^7) $$
$$ r = 4.215 \times 10^7 \text{ m} $$
Now, if the mass of Earth is four times its original mass:
$$ M' = 4M $$
The new time period will be:
$$ T' = 2\pi \sqrt{\frac{r^3}{G(4M)}} $$
Since:
$$ T = 2\pi \sqrt{\frac{r^3}{GM}} $$
Dividing both equations:
$$ \frac{T'}{T} = \sqrt{\frac{1}{4}} $$
$$ T' = \frac{T}{2} $$
For a geostationary satellite, the normal time period is 24 hours. So,
$$ T' = \frac{24}{2} = 12 \text{ hours} $$
Therefore, the satellite would take 12 hours to complete one orbit.
Solution 2: Simple and Understandable Answer
Satellites orbit Earth based on gravity. The formula for the time it takes is:
$$ T = 2\pi \sqrt{\frac{r^3}{GM}} $$
If Earth's mass becomes 4 times bigger, gravity increases, and satellites move faster.
Using the formula, we find that the time period will become half of the original time.
A geostationary satellite normally takes 24 hours. Since gravity is stronger in this case, it will complete its orbit in:
12 hours
So, the satellite will take 12 hours to go around Earth!
Solve the problem.
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 2√2T seconds for one revolution?
Solution 1: Scientific and Written Exam Answer
The time period of a satellite in orbit is given by Kepler’s third law:
$$ T \propto r^{\frac{3}{2}} $$
Where:
- T = Time period of the satellite
- r = Distance from the center of the Earth
Let the radius of the first satellite be:
$$ r_1 = R + h_1 $$
For the second satellite, the time period is increased to 2√2T, so:
$$ T_2 = 2\sqrt{2} T_1 $$
Using Kepler’s law:
$$ \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{\frac{3}{2}} $$
Substituting the values:
$$ 2\sqrt{2} = \left(\frac{r_2}{r_1}\right)^{\frac{3}{2}} $$
Taking the power of $$ \frac{2}{3} $$ on both sides:
$$ \left(2\sqrt{2}\right)^{\frac{2}{3}} = \frac{r_2}{r_1} $$
Solving:
$$ 2^{\frac{2}{3}} \times 2^{\frac{1}{3}} = 2^{\frac{3}{3}} = 2 $$
So:
$$ r_2 = 2 r_1 $$
Now, since:
$$ r_1 = R + h_1 $$
We get:
$$ R + h_2 = 2(R + h_1) $$
Solving for $$ h_2 $$:
$$ h_2 = 2(R + h_1) - R $$
$$ h_2 = 2h_1 + R $$
Therefore, the new height of the satellite is:
$$ h_2 = 2h_1 + R $$
Solution 2: Simple and Understandable Answer
Kepler’s third law tells us that a satellite’s orbit time depends on its distance from the center of the Earth.
If one satellite completes its orbit in T seconds at a height h₁, and another satellite takes 2√2T seconds, its height will be higher.
The formula says:
$$ h_2 = 2h_1 + R $$
That means the second satellite is twice as high as the first one, plus the radius of Earth!
So, the second satellite’s height will be 2 times the first height plus Earth’s radius.