Gravitation Questions and Answers Class 10th Science | Solutions for Chapter 1: Gravitation
Study the entries in the following table and rewrite them, putting the connected items in a single row.
Question Table:
| I | II | III |
|---|---|---|
| Mass | m/s² | Zero at the centre |
| Weight | kg | Measure of inertia |
| Acceleration due to gravity | Nm²/kg² | Same in the entire universe |
| Gravitational constant | N | Depends on height |
Corrected Table:
| I | II | III |
|---|---|---|
| Mass | kg | Measure of inertia |
| Weight | N | Zero at the centre |
| Acceleration due to gravity | m/s² | Depends on height |
| Gravitational constant | Nm²/kg² | Same in the entire universe |
What is the difference between mass and weight of an object?
Comparison of Mass and Weight:
| Property | Mass | Weight |
|---|---|---|
| Definition | Mass is the amount of matter present in an object. | Weight is the force exerted by gravity on an object. |
| Unit | Measured in kilograms (kg) or grams (g). | Measured in newtons (N) or pounds (lb). |
| Nature | Scalar quantity (only magnitude). | Vector quantity (magnitude and direction). |
| Dependence | Does not depend on location. | Depends on gravitational field and changes with location. |
| Constancy | Remains constant everywhere. | Changes depending on gravitational pull. |
| Measurement Tool | Measured using a balance. | Measured using a spring balance or weighing scale. |
| Effect of Gravity | Not affected by gravity. | Directly affected by gravity. |
| Example | A 10 kg object on Earth, the Moon, or in space has the same mass. | A 10 kg object on Earth weighs about 98 N, but on the Moon, it weighs only 16.5 N. |
| Formula | Mass = Density × Volume | Weight = Mass × Gravitational Force (W = mg) |
| Relation with Inertia | Mass represents the inertia of an object. | Weight has no direct relation to inertia. |
Will the mass and weight of an object on the earth be same as their values on Mars? Why?
Scientific and Written Exam Answer:
Mass is a fundamental property of an object and remains constant irrespective of location. Therefore, the mass of an object on Earth and Mars will be the same.
However, weight is the force exerted by gravity on an object and is given by the formula:
$$ W = mg $$
where,
W = Weight of the object (N)
m = Mass of the object (kg)
g = Acceleration due to gravity (m/s²)
The gravitational acceleration on Mars (3.7 m/s²) is lower than on Earth (9.8 m/s²). Since weight depends on gravity, an object will weigh less on Mars than on Earth.
Simple and Understandable Answer:
No, the mass and weight of an object will not be the same on Earth and Mars.
Mass is a fixed property and does not change. If an object has a mass of 10 kg on Earth, it will also have a mass of 10 kg on Mars.
Weight, however, depends on gravity. Since Mars has weaker gravity than Earth, an object will weigh less on Mars. For example, if a person weighs 60 kg on Earth, they will weigh much less on Mars.
What is free fall?
Scientific and Written Exam Answer:
Free fall is the motion of an object under the influence of gravitational force only, without any resistance from air or other external forces.
In free fall, the acceleration of the object is equal to the acceleration due to gravity (g), and the equation of motion can be applied as:
$$ s = ut + \frac{1}{2}gt^2 $$
where,
s = Distance fallen (m)
u = Initial velocity (m/s)
t = Time taken (s)
g = Acceleration due to gravity (9.8 m/s² on Earth)
Since free-falling objects experience no other forces except gravity, they appear weightless.
Simple and Understandable Answer:
Free fall happens when an object falls only due to gravity, without any push, pull, or air resistance.
For example, if you drop a ball from a tall building, and there is no air resistance, it will fall freely under gravity. This is called free fall.
Astronauts in space experience free fall, which makes them feel weightless.
What is the acceleration due to gravity?
Scientific and Written Exam Answer:
The acceleration due to gravity (g) is the acceleration experienced by an object when it is in free fall under the influence of the Earth's gravitational force.
Its standard value on Earth's surface is:
$$ g = 9.8 \, m/s^2 $$
It is derived using Newton's universal law of gravitation:
$$ g = \frac{GM}{R^2} $$
where,
G = Universal gravitational constant (6.674 × 10⁻¹¹ Nm²/kg²)
M = Mass of Earth (5.972 × 10²⁴ kg)
R = Radius of Earth (~6371 km)
The value of g decreases with altitude and varies slightly based on location.
Simple and Understandable Answer:
Acceleration due to gravity (g) is the force that pulls objects toward the Earth when they fall.
Its value on Earth is 9.8 m/s². This means that when an object falls freely, its speed increases by 9.8 m/s every second.
For example, if you drop a ball, it will fall faster as it moves downward because of gravity.
What is escape velocity?
Scientific and Written Exam Answer:
Escape velocity is the minimum velocity required for an object to escape a planet’s gravitational pull without any further propulsion.
The formula for escape velocity is:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
where,
vₑ = Escape velocity (m/s)
G = Universal gravitational constant (6.674 × 10⁻¹¹ Nm²/kg²)
M = Mass of the planet (kg)
R = Radius of the planet (m)
For Earth, the escape velocity is approximately:
$$ v_e = 11.2 \, km/s $$
This means an object must travel at 11.2 km/s to break free from Earth's gravity.
Simple and Understandable Answer:
Escape velocity is the speed an object needs to leave a planet’s gravity without falling back.
For Earth, this speed is 11.2 km/s. If a rocket moves slower than this, gravity will pull it back. But if it reaches or exceeds this speed, it can travel into space.
For example, spacecraft leaving Earth must reach this speed to go beyond our atmosphere.
What is centripetal force?
Scientific and Written Exam Answer:
Centripetal force is the force that acts on an object moving in a circular path, directing it toward the center of the circle.
The formula for centripetal force is:
$$ F_c = \frac{m v^2}{r} $$
where,
Fₐ = Centripetal force (N)
m = Mass of the object (kg)
v = Velocity of the object (m/s)
r = Radius of the circular path (m)
Without centripetal force, objects would move in a straight line due to inertia.
Simple and Understandable Answer:
Centripetal force is the force that keeps an object moving in a circle instead of flying off in a straight line.
For example, when you swing a stone tied to a string, the tension in the string acts as centripetal force, pulling the stone towards the center.
If the string breaks, the stone moves in a straight line instead of continuing in a circle.
Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?
Scientific and Written Exam Answer:
Kepler's Three Laws of Planetary Motion:
1. Law of Ellipses: Every planet moves around the Sun in an elliptical orbit, with the Sun at one of the two foci.
2. Law of Equal Areas: A line segment joining a planet and the Sun sweeps out equal areas in equal intervals of time.
3. Law of Harmonies: The square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit.
Mathematically,
$$ T^2 \propto r^3 $$
Kepler’s Laws and Newton’s Inverse Square Law:
Kepler’s laws helped Newton derive the law of universal gravitation:
1. The elliptical motion of planets (Kepler's First Law) suggested that a central force must be acting towards the Sun.
2. The Law of Equal Areas (Kepler's Second Law) indicated that this force should always act towards the center.
3. The Law of Harmonies (Kepler's Third Law) led Newton to establish the relation:
$$ F \propto \frac{1}{r^2} $$
which is the inverse square law of gravity:
$$ F = \frac{G M m}{r^2} $$
where:
F = Gravitational force (N)
G = Universal gravitational constant (6.674 × 10⁻¹¹ Nm²/kg²)
M = Mass of the larger body (kg)
m = Mass of the smaller body (kg)
r = Distance between the two masses (m)
Simple and Understandable Answer:
Kepler's Three Laws:
1. Planets move in elliptical orbits: Instead of moving in perfect circles, planets travel around the Sun in an oval-shaped path.
2. Planets move faster when closer to the Sun: A planet covers the same area in the same amount of time, meaning it moves quicker when it is near the Sun and slower when it is farther.
3. The farther a planet is from the Sun, the longer it takes to complete one orbit: Planets that are farther away take more time to revolve around the Sun.
How These Laws Helped Newton:
Kepler’s laws helped Newton understand that the force keeping planets in orbit must decrease as the distance increases.
This led him to discover the inverse square law of gravity, which means that the force of attraction between two objects reduces as the square of the distance between them increases.
Example: If the distance between two objects is doubled, the gravitational force between them becomes four times weaker.
Gravitation Class 10 Questions And Answers Maharashtra Board
A stone is thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is the same as the time taken to come down.
Scientific and Written Exam Answer:
To prove that the time taken to go up is equal to the time taken to come down, we use the equations of motion.
Given:
Initial velocity = u (upward)
Acceleration due to gravity = g (acting downward)
Final velocity at the highest point = 0 (velocity becomes zero at maximum height)
Using the first equation of motion for upward motion:
$$ v = u - g t_1 $$
At maximum height, \( v = 0 \), so:
$$ 0 = u - g t_1 $$
Solving for \( t_1 \):
$$ t_1 = \frac{u}{g} $$
Now, for downward motion:
Initial velocity at the top = 0
Acceleration due to gravity = g (acting downward)
Final velocity when reaching the ground = u
Using the first equation of motion for downward motion:
$$ v = u + g t_2 $$
Since \( v = u \), we get:
$$ u = 0 + g t_2 $$
Solving for \( t_2 \):
$$ t_2 = \frac{u}{g} $$
Since \( t_1 = t_2 \), we conclude:
The time taken to go up is equal to the time taken to come down.
Simple and Understandable Answer:
When a stone is thrown upwards, it slows down due to gravity until it reaches the highest point where its velocity becomes zero. Then, it starts coming back down with increasing speed due to gravity.
The same gravitational force that slows it down while going up also speeds it up while coming down. As a result, the time taken to rise to the highest point is equal to the time taken to fall back to the starting position.
Example: If a ball is thrown up and takes 3 seconds to reach the top, it will also take 3 seconds to come back down.
If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?
Scientific and Written Exam Answer:
The force required to pull an object depends on the frictional force between the object and the floor. Friction is directly proportional to the normal force, which is equal to the weight of the object. The weight of an object is given by:
$$ W = mg $$
where:
- \( W \) = Weight of the object
- \( m \) = Mass of the object
- \( g \) = Acceleration due to gravity
If the value of \( g \) doubles, the new weight becomes:
$$ W' = m(2g) = 2mg $$
Since the normal force is equal to the weight of the object, it also doubles. The frictional force is given by:
$$ f = \mu N $$
where:
- \( f \) = Frictional force
- \( \mu \) = Coefficient of friction
- \( N \) = Normal force
As \( N \) doubles, the frictional force \( f \) also doubles. This means pulling the object requires twice the force, making it two times more difficult to move.
Simple and Understandable Answer:
The weight of an object depends on gravity. If gravity becomes twice as strong, the object will feel twice as heavy.
Since friction depends on the weight of the object, more weight means more friction. This extra friction makes it harder to pull or push the object along the floor.
Example: Imagine trying to slide a box on a smooth floor. If the box suddenly becomes twice as heavy, you would need to use twice as much effort to move it.
Explain why the value of g is zero at the centre of the earth.
Scientific and Written Exam Answer:
The acceleration due to gravity (\( g \)) at a distance \( r \) from the center of the Earth inside its surface is given by:
$$ g' = g \left( \frac{r}{R} \right) $$
where:
- \( g' \) = Acceleration due to gravity at depth \( r \)
- \( g \) = Acceleration due to gravity at the surface of the Earth
- \( R \) = Radius of the Earth
At the center of the Earth, \( r = 0 \), so:
$$ g' = g \left( \frac{0}{R} \right) = 0 $$
This means that the value of \( g \) at the center of the Earth is zero. The reason is that at the center, the mass of the Earth is symmetrically distributed in all directions, causing the gravitational forces to cancel out, resulting in no net gravitational pull.
Simple and Understandable Answer:
Gravity pulls everything towards the center of the Earth. However, at the exact center, the pull from all sides is equal and cancels out. Since there is no force pulling in any specific direction, gravity becomes zero at that point.
Example: Imagine standing in the middle of a room with people pulling you equally from all directions. Since every pull is balanced, you do not move. Similarly, at the Earth’s center, gravity cancels out, making \( g = 0 \).
Maharashtra Solutions for Science and Technology 1 [English] 10 Standard SSC Chapter 1 Gravitation
Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be √8T.
Scientific and Written Exam Answer:
According to Kepler's Third Law, the square of the time period (\(T\)) of a planet is directly proportional to the cube of its orbital radius (\(R\)):
$$ T^2 \propto R^3 $$
This means:
$$ \frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} $$
Given:
- Initial distance: \( R_1 = R \)
- New distance: \( R_2 = 2R \)
- Initial period: \( T_1 = T \)
- New period: \( T_2 = ? \)
Substituting these values:
$$ \frac{T_2^2}{T^2} = \frac{(2R)^3}{R^3} $$
$$ \frac{T_2^2}{T^2} = \frac{8R^3}{R^3} = 8 $$
Taking square root on both sides:
$$ T_2 = \sqrt{8} T = 2\sqrt{2} T $$
Thus, if the planet moves to a distance of \( 2R \), its new period of revolution will be \( 2\sqrt{2}T \).
Simple and Understandable Answer:
Kepler’s third law tells us that if a planet moves farther from a star, it takes longer to complete its orbit. The time period (\( T \)) increases with the distance.
If the planet moves from \( R \) to \( 2R \), its new time period becomes \( 2\sqrt{2}T \), which is approximately **2.83 times the original time**.
Example: Imagine running around a circular track. If the track becomes bigger, you will take more time to complete one round. Similarly, as the planet moves farther, its orbit becomes larger, and its revolution takes more time.
An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?
Scientific and Written Exam Answer:
To find the value of acceleration due to gravity (\( g \)), we use the equation of motion:
$$ s = ut + \frac{1}{2} g t^2 $$
Given:
- Initial velocity, \( u = 0 \) (since the object is dropped, not thrown)
- Displacement, \( s = 5 \) m
- Time, \( t = 5 \) s
Substituting these values into the equation:
$$ 5 = 0 \times 5 + \frac{1}{2} g (5)^2 $$
$$ 5 = \frac{1}{2} g \times 25 $$
Multiplying both sides by 2:
$$ 10 = 25g $$
Solving for \( g \):
$$ g = \frac{10}{25} $$
$$ g = 0.4 \, \text{m/s}^2 $$
Thus, the acceleration due to gravity on this planet is \( 0.4 \, \text{m/s}^2 \).
Simple and Understandable Answer:
We can find the gravity of the planet using a simple formula that relates height, time, and gravity.
The formula is:
$$ s = ut + \frac{1}{2} g t^2 $$
Since the object starts from rest (\( u = 0 \)), the equation simplifies. After plugging in the given values, we calculate \( g \) and find that it is \( 0.4 \, \text{m/s}^2 \).
Example: Imagine dropping a ball from a table. If the planet has less gravity than Earth, the ball will fall slowly. Here, the object takes 5 seconds to fall just 5 meters, which means gravity is much weaker than on Earth.
The radius of planet A is half the radius of planet B. If the mass of A is \( M_A \), what must be the mass of B so that the value of \( g \) on B is half that of its value on A?
Scientific and Written Exam Answer:
The acceleration due to gravity \( g \) on a planet is given by the formula:
$$ g = \frac{G M}{R^2} $$
Where:
- \( G \) = Gravitational constant
- \( M \) = Mass of the planet
- \( R \) = Radius of the planet
For planet A:
$$ g_A = \frac{G M_A}{R_A^2} $$
For planet B:
$$ g_B = \frac{G M_B}{R_B^2} $$
Given that the radius of planet A is half the radius of planet B:
$$ R_A = \frac{1}{2} R_B $$
Also, it is given that \( g_B \) is half of \( g_A \):
$$ g_B = \frac{1}{2} g_A $$
Substituting the values:
$$ \frac{G M_B}{R_B^2} = \frac{1}{2} \times \frac{G M_A}{R_A^2} $$
Replacing \( R_A^2 \) with \( \left( \frac{1}{2} R_B \right)^2 \):
$$ \frac{G M_B}{R_B^2} = \frac{1}{2} \times \frac{G M_A}{\left( \frac{1}{2} R_B \right)^2} $$
$$ \frac{G M_B}{R_B^2} = \frac{1}{2} \times \frac{G M_A}{\frac{1}{4} R_B^2} $$
$$ \frac{G M_B}{R_B^2} = \frac{1}{2} \times \frac{4 G M_A}{R_B^2} $$
$$ M_B = 2 M_A $$
Thus, the mass of planet B must be twice the mass of planet A (\( M_B = 2 M_A \)) for its gravitational acceleration to be half that of planet A.
Simple and Understandable Answer:
The acceleration due to gravity \( g \) depends on a planet's mass and radius according to the formula:
$$ g = \frac{G M}{R^2} $$
Since planet B is bigger (its radius is twice that of planet A), its gravity will decrease. To balance this, its mass must be adjusted. After calculation, we find that the mass of planet B must be twice that of planet A to make gravity half as strong.
Example: Imagine Earth and another planet with double its size. If we want the new planet to have only half the gravity of Earth, it must have twice Earth's mass to maintain the correct balance.
The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.
Scientific and Written Exam Answer:
The mass of an object remains the same everywhere in the universe. Therefore, the mass of the object on the Moon will be:
$$ m = 5 \text{ kg} $$
Weight is calculated using the formula:
$$ W = mg $$
On Earth:
$$ W_E = 5 \times 9.8 $$
$$ W_E = 49 \text{ N} $$
On the Moon, gravity is \( \frac{1}{6} \)th of that on Earth:
$$ g_M = \frac{9.8}{6} = 1.63 \text{ m/s}^2 $$
So, the weight on the Moon will be:
$$ W_M = 5 \times 1.63 $$
$$ W_M = 8.17 \text{ N} $$
Thus, on the Moon:
- Mass = 5 kg (remains the same)
- Weight = 8.17 N
Simple and Understandable Answer:
Mass is a measure of how much matter is in an object, and it does not change no matter where the object is. So, the mass of the object on the Moon will still be 5 kg.
Weight depends on gravity. Since the Moon’s gravity is much weaker (1/6th of Earth’s gravity), the object will weigh much less.
On Earth, the object weighs 49 N. Since the Moon’s gravity is 6 times weaker, the object’s weight on the Moon will be:
$$ W_M = \frac{49}{6} = 8.17 \text{ N} $$
Example: Imagine carrying a heavy bag on Earth and then carrying it on the Moon. The bag would feel much lighter on the Moon because the pull of gravity is weaker!
Balbharati solutions for Science and Technology 1 [English] 10 Standard SSC Maharashtra State Board chapter 1 - Gravitation [Latest edition]
An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s².
Scientific and Written Exam Answer:
We use the equation of motion:
$$ v^2 = u^2 - 2gh $$
At the highest point, the final velocity \( v = 0 \).
Given:
- Maximum height, \( h = 500 \) m
- Acceleration due to gravity, \( g = 10 \) m/s²
- Final velocity, \( v = 0 \)
Substituting the values:
$$ 0 = u^2 - 2(10)(500) $$
$$ u^2 = 10000 $$
$$ u = 100 \text{ m/s} $$
Time to reach the highest point:
Using the formula:
$$ v = u - gt $$
$$ 0 = 100 - (10 \times t) $$
$$ t = 10 \text{ s} $$
Total time for the object to return to Earth:
Since the time taken to go up is equal to the time taken to come down,
$$ T = 2 \times 10 = 20 \text{ s} $$
Final Answer:
- Initial velocity (u) = 100 m/s
- Total time to return = 20 s
Simple and Understandable Answer:
When an object is thrown upwards, it slows down due to gravity until it stops at the highest point. The speed at which it was thrown is called the initial velocity.
We used a formula to calculate it and found that the object was thrown with an initial velocity of 100 m/s.
Next, we calculated how long the object takes to reach the highest point. It takes 10 seconds to go up, and since falling down takes the same time, the total time to come back is 20 seconds.
Example: Imagine you throw a ball straight up. The stronger you throw it, the higher it goes. When it reaches the top, it stops for a moment and then comes back down. Here, we calculated how fast it was thrown and how long it stays in the air!
A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s², calculate its speed on reaching the ground and the height of the table.
Scientific and Written Exam Answer:
We use the equations of motion to solve this problem.
1. Speed on reaching the ground:
Using the equation:
$$ v = u + gt $$
Given:
- Initial velocity, \( u = 0 \) (since the ball falls from rest)
- Acceleration due to gravity, \( g = 10 \) m/s²
- Time, \( t = 1 \) s
Substituting the values:
$$ v = 0 + (10 \times 1) $$
$$ v = 10 \text{ m/s} $$
2. Height of the table:
Using the equation:
$$ h = ut + \frac{1}{2} g t^2 $$
Substituting the values:
$$ h = (0 \times 1) + \frac{1}{2} (10) (1^2) $$
$$ h = 5 \text{ m} $$
Final Answer:
- Speed on reaching the ground = 10 m/s
- Height of the table = 5 m
Simple and Understandable Answer:
When a ball falls off a table, gravity pulls it down, making it speed up. The longer it falls, the faster it moves.
We used a formula to find that by the time it hits the ground, its speed is 10 m/s.
We also calculated the height of the table using another formula and found it to be 5 meters.
Example: Imagine dropping a coin from a table. It starts from rest and falls faster as it reaches the ground. Here, we calculated its speed just before hitting the ground and the height from which it fell!
The masses of the earth and moon are 6 × 10²⁴ kg and 7.4 × 10²² kg, respectively. The distance between them is 3.8 × 10⁵ km. Calculate the gravitational force of attraction between the two?
(Use G = 6.7 × 10⁻¹¹ N·m²/kg²)
Scientific and Written Exam Answer:
We use Newton’s Universal Law of Gravitation:
$$ F = \frac{G M_1 M_2}{r^2} $$
Where:
- \( F \) = Gravitational force
- \( G = 6.7 \times 10^{-11} \) N·m²/kg²
- \( M_1 = 6 \times 10^{24} \) kg (Mass of Earth)
- \( M_2 = 7.4 \times 10^{22} \) kg (Mass of Moon)
- \( r = 3.8 \times 10^{5} \) km = \( 3.8 \times 10^{8} \) m (converted to meters)
Step 1: Substituting the values in the formula:
$$ F = \frac{(6.7 \times 10^{-11}) (6 \times 10^{24}) (7.4 \times 10^{22})}{(3.8 \times 10^8)^2} $$
Step 2: Squaring the distance:
$$ (3.8 \times 10^8)^2 = 14.44 \times 10^{16} $$
Step 3: Multiplying the masses:
$$ (6 \times 10^{24}) \times (7.4 \times 10^{22}) = 44.4 \times 10^{46} $$
Step 4: Applying the values:
$$ F = \frac{(6.7 \times 10^{-11}) \times (44.4 \times 10^{46})}{14.44 \times 10^{16}} $$
Step 5: Calculating the numerator:
$$ (6.7 \times 44.4) = 297.48 $$
$$ 297.48 \times 10^{35} $$
Step 6: Final Calculation:
$$ F = \frac{297.48 \times 10^{35}}{14.44 \times 10^{16}} $$
$$ F = 20.6 \times 10^{19} $$
$$ F = 2.06 \times 10^{20} \text{ N} $$
Final Answer: 2.06 × 10²⁰ N
Simple and Understandable Answer:
The Earth and Moon attract each other due to gravity. This force is given by the formula:
$$ F = \frac{G M_1 M_2}{r^2} $$
By plugging in the values, we found that the gravitational force between Earth and the Moon is 2.06 × 10²⁰ N.
Example: If you bring two magnets closer, they pull each other. Similarly, the Earth and Moon pull each other due to gravity!
Balbharati solutions for Science and Technology 1 [English] 10 Standard SSC Maharashtra State Board 1 Gravitation Exercises [Pages 14 - 15]
The mass of the earth is 6 × 10²⁴ kg. The distance between the earth and the sun is 1.5 × 10¹¹ m. If the gravitational force between the two is 3.5 × 10²² N, what is the mass of the sun?
(Use G = 6.7 × 10⁻¹¹ N·m²/kg²)
Step-by-Step Solution:
We use Newton’s Universal Law of Gravitation:
$$ F = \frac{G M_1 M_2}{r^2} $$
Where:
- \( F = 3.5 \times 10^{22} \) N (Gravitational force)
- \( G = 6.7 \times 10^{-11} \) N·m²/kg²
- \( M_1 = 6 \times 10^{24} \) kg (Mass of Earth)
- \( M_2 \) = ? (Mass of Sun - to be found)
- \( r = 1.5 \times 10^{11} \) m (Distance between Earth and Sun)
Step 1: Rearranging the formula to find \( M_2 \):
$$ M_2 = \frac{F \times r^2}{G \times M_1} $$
Step 2: Substituting the values:
$$ M_2 = \frac{(3.5 \times 10^{22}) \times (1.5 \times 10^{11})^2}{(6.7 \times 10^{-11}) \times (6 \times 10^{24})} $$
Step 3: Squaring the distance:
$$ (1.5 \times 10^{11})^2 = 2.25 \times 10^{22} $$
Step 4: Multiplying the numerator:
$$ (3.5 \times 10^{22}) \times (2.25 \times 10^{22}) = 7.875 \times 10^{44} $$
Step 5: Multiplying the denominator:
$$ (6.7 \times 10^{-11}) \times (6 \times 10^{24}) = 4.02 \times 10^{14} $$
Step 6: Final Calculation:
$$ M_2 = \frac{7.875 \times 10^{44}}{4.02 \times 10^{14}} $$
$$ M_2 = 1.96 \times 10^{30} $$
Final Answer: 1.96 × 10³⁰ kg
Simple Explanation:
The Earth and the Sun attract each other due to gravity. Using the given values and the formula for gravitational force, we calculated that the mass of the Sun is 1.96 × 10³⁰ kg.
Example: The Sun is extremely massive, which is why it has a strong gravitational pull on all the planets in the solar system!