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Gravitation Questions and Answers 10 th class science ch1 part 1

 Q1. Study the entries in the following table and rewrite them putting the connected items in a single row.

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

 Solution:-

I II III
Mass kg Measure of inertia
Weight N Depends on height
Acceleration due to gravity m/s2 Zero at the center
Gravitational constant Nm2/kg2 Same in the entire universe


Q2. 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?

Answer:- To calculate the value of acceleration due to gravity (g) on the planet, we can use the equation of motion for free fall:

h = (1/2) x g x t2

Where:

h = height

g = acceleration due to gravity

t = time

Given:

Height (h) = 5 m

Time (t) = 5 s

Plugging in the values, we have:

5 = (1/2) x g x (52)

5 = (1/2) x g x 25

5 = 12.5g

To find the value of g, we can rearrange the equation:

g = 5 / 12.5

g = 0.4 m/s2

Therefore, the value of acceleration due to gravity (g) on the planet is 0.4 m/s2.


Q3. What is the difference between mass and weight of an object. Will the mass and weight of an object on the earth be same as their values on Mars? Why?

Answer:- The difference between mass and weight of an object lies in their definitions.

        1.Mass: Mass is a measure of the amount of matter contained in an object. It is a fundamental property and is independent of the object's location. Mass is typically measured in kilograms (kg).

        2. Weight: Weight is the force exerted on an object due to gravity. It is the gravitational force acting on an object's mass. Weight depends on the mass of the object and the strength of the gravitational field it is in. Weight is typically measured in newtons (N).

    On Mars, the acceleration due to gravity is weaker than on Earth, approximately 3.7 m/s². This means that the same object would weigh less on Mars compared to its weight on Earth. So, the mass of an object remains constant regardless of its location, but the weight can vary depending on the strength of the gravitational field.


Q4.What are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force ? 
Answer:-     (A) Free fall:- A free fall is the motion of an object falling solely due to gravity, without the assistance of any other forces. An object in free fall experiences an acceleration from gravity, and as it falls, its velocity rises steadily. The only force acting on an object in free fall is gravity, which causes it to fall faster and faster towards the Earth.
        (B) Acceleration due to gravity: Acceleration due to gravity, denoted as "g," is the acceleration experienced by an object due to the gravitational force of the Earth (or any other celestial body). It represents the rate at which an object's velocity changes under the influence of gravity. On the surface of the Earth, the average value of acceleration due to gravity is approximately 9.8 m/s². This means that in free fall, the velocity of an object increases by 9.8 meters per second every second.
        (C) Escape velocity: Escape velocity is the minimum velocity an object must achieve to escape the gravitational pull of a celestial body. It is the speed required to overcome the gravitational force and move away from the object indefinitely. The escape velocity depends on the mass and radius of the celestial body. For example, the escape velocity on Earth is approximately 11.2 km/s. If an object is launched with a velocity greater than or equal to the escape velocity, it will escape Earth's gravitational field.
        (D) Centripetal force: Centripetal force is the force that acts on an object moving in a circular path, directing it toward the center of the circle. It is necessary to keep the object in its circular motion. The centripetal force is always directed inward, while the object's velocity is directed tangentially to the circular path. The magnitude of the centripetal force depends on the mass of the object, its speed, and the radius of the circular path. It can be provided by various forces such as tension in a string, gravitational force, or friction.

Q5. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?
Answer:- 1. Kepler's First Law (Law of Ellipses): Each planet orbits the Sun in an elliptical path, with the Sun at one of the two foci of the ellipse.
        2. Kepler's Second Law (Law of Equal Areas): A line connecting a planet to the Sun sweeps out equal areas in equal time intervals. This means that a planet moves faster when it is closer to the Sun and slower when it is farther away.
        3. Kepler's Third Law (Harmonic Law): The square of the orbital period of a planet is directly proportional to the cube of its average distance from the Sun. Mathematically, it can be expressed as T² = k × r³, where T represents the period of revolution, r represents the average distance from the Sun, and k is a constant.
        Newton built upon Kepler's laws and used them as a foundation for his theory of universal gravitation. By combining Kepler's laws with his laws of motion, Newton was able to derive the inverse square law of gravity. He showed that the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This relationship is described by the equation F = G × (m1 × m2) / r², where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.

        Kepler's laws provided Newton with observational data and patterns in planetary motion, which helped him formulate the concept of a gravitational force acting between celestial bodies. By combining these laws with his own principles of motion, Newton developed a comprehensive theory of gravity that explained the motion of not only planets but also other objects on Earth and in the universe

Q.6 A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is same as the time taken to come down.
Answer:- We know that the upward motion and downward motion of the stone are symmetrical, meaning they follow the same path but in opposite directions. Therefore, the time taken to reach the maximum height during the upward motion will be equal to the time taken to descend from the maximum height during the downward motion.

During the upward motion:
Initial velocity (u) = u
Final velocity (v) = 0 (at the highest point)
Acceleration (a) = acceleration due to gravity (g) = -9.8 m/s2 (negative because it acts in the opposite direction of motion)

Using the equation of motion: v = u + at
0 = u - 9.8t₁   ------ (1)

During the downward motion:
Initial velocity (u') = 0 (at the highest point)
Final velocity (v') = -u (negative because it is in the opposite direction)
Acceleration (a) = acceleration due to gravity (g) = -9.8 m/s^2

Using the equation of motion: v' = u' + at'
-u = -9.8t₂   ------ (2)

From equation (1), we have: 0 = u - 9.8t₁
Solving for t₁: t₁ = u/9.8

From equation (2), we have: -u = -9.8t₂
Solving for t₂: t₂ = u/9.8

As we can see, the time taken to go up (t₁) is equal to the time taken to come down (t₂). Therefore, the time taken to go up is the same as the time taken to come down for a stone thrown vertically upwards.

Q7. 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?
Answer:- The value of g represents the acceleration due to gravity, which determines the force with which objects are pulled towards the Earth's surface. If the value of g suddenly becomes twice its original value, it means that the gravitational force acting on objects has increased.
        If the value of g becomes twice its original value, the weight of the object will also double. As a result, the force required to pull the heavy object along the floor will need to be increased. This is because the increased gravitational force makes it more difficult to overcome the friction and move the object.

Q8. Explain why the value of g is zero at the centre of the earth.
Answer:- The value of g represents the acceleration due to gravity, which is the gravitational force experienced by an object near the surface of a planet. The value of g is not zero at the center of the Earth but rather decreases as you move towards the center.
        At the center of the Earth, the gravitational forces from all directions cancel each other out, resulting in a net gravitational force of zero. This means that objects at the center of the Earth experience no gravitational acceleration and feel weightless.

Q9. 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 square root √8 T. 
Answer:- According to Kepler's third law, the square of the period of revolution (T) of a planet is directly proportional to the cube of its semi-major axis (a).

T² ∝ a³

In the given scenario, let the period of revolution at distance R be T. We need to find the period of revolution (T') when the planet is at a distance of 2R from the star.

Using Kepler's third law, we have:

T² = kR³ ......(1)

where k is a constant of proportionality.

When the distance is 2R, we can express the period of revolution as:

T'² = k(2R)³
T'² = 8kR³

Taking the ratio of equation (1) and (2):

(T'² / T²) = (8kR³) / (kR³)
(T'² / T²) = 8

Taking the square root of both sides:

T' / T = √8
T' / T = 2√2

Therefore, the period of revolution T' at a distance of 2R from the star is equal to 2 times the square root of 2 times the original period T.

Q10. The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?
Answer:- Let's assume the radius of planet A is denoted by RA and the radius of planet B is denoted by RB. Given that the radius of planet A is half the radius of planet B, we can express this as:

RA = (1/2)RB

Now, we need to find the mass of planet B (MB) such that the value of acceleration due to gravity (g) on planet B is half that of its value on planet A.

The acceleration due to gravity is given by:

g = (G * M) / R²

where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.

For planet A, we have:

gA = (G * MA) / RA²

For planet B, we have:

gB = (G * MB) / RB²

Given that gB is half of gA, we can write:

gB = (1/2)gA

Substituting the expressions for gA and gB, we get:

(G * MB) / RB² = (1/2) * [(G * MA) / RA²]

Simplifying the equation, we can cancel out the gravitational constant G:

MB / RB² = (1/2) * (MA / RA²)

Now, substituting the relationship between RA and RB, we have:

MB / RB² = (1/2) * (MA / (1/2)²RB²)
MB / RB² = (1/2) * (4MA / RB²)
MB / RB² = 2MA / RB²
MB = 2MA

Therefore, the mass of planet B (MB) must be twice the mass of planet A (MA) in order for the value of g on B to be half that of its value on A.

Q11. 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. 
Answer:- On the Moon, the acceleration due to gravity is 1/6th of that on Earth. 

Given:
Mass on Earth (MA) = 5 kg
Weight on Earth (WA) = 49 N

To find the values on the Moon, we can use the relation:

Weight = Mass x Acceleration due to gravity

Since the mass remains the same, we can calculate the weight on the Moon as follows:

Weight on Moon (WM) = Mass x Acceleration due to gravity on Moon

Acceleration due to gravity on Moon = (1/6) x Acceleration due to gravity on Earth

WM = MA x (1/6) x WA

Substituting the given values:

WM = 5 kg x (1/6) x 49 N
WM = 8.17 N

Therefore, the mass of the object on the Moon remains 5 kg, and its weight on the Moon is approximately 8.17 N.

Q12.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
Answer:- To find the initial velocity of the object, we can use the equation of motion:

v2 = u2 + 2as

Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (to be determined)
a = acceleration due to gravity (g = 10 m/s2
s = displacement (500 m upwards)

At the highest point, the final velocity is 0 m/s, so the equation becomes:

0 = u2

0 = u2

u2

Taking the square root of both sides:

u = ±√10000

u = ±100 m/s

Since the object is thrown vertically upwards, we take the negative value:

u = -100 m/s

Therefore, the initial velocity of the object is -100 m/s (indicating upward direction).

To find the time taken to come back to Earth, we can use the equation of motion:

v = u + at

Where:
v = final velocity (0 m/s at the starting point)
u = initial velocity (-100 m/s)
a = acceleration due to gravity (g = 10 m/s2)
t = time taken to reach the starting point (to be determined)

0 = -100 + 10t

10t = 100

t = 10 s

This is only the time taken for the upward journey. To find the total time taken to come back to Earth, we need to consider the downward journey as well. The downward journey will take the same amount of time as the upward journey. Therefore, the total time taken to come back to Earth is:

Total time = 10 s (upward journey) + 10 s (downward journey)

Total time = 20 s

Therefore, the object will take 20 seconds to come back to Earth.


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