**1. Two objects are released simultaneously from different heights. If one object takes 2 seconds to reach the ground, how much time will the other object take to reach the ground if its height is four times the height of the first object?**

a) 4 s

b) 2 s

c) 8 s

d) 1 s

**Answer:** a) To solve this question we can use the following equation for free-falling objects:

h = ut + 1/2 g t^{2}

Since the initial velocity u is zero, the equation simplifies to:

h = 1/2 g t^{2}

To find the time t, we rearrange the equation as follows:

Where:

t is the time of fall,

h is the height from which the object is dropped,

g is the acceleration due to gravity (approximated as 9.8 ms^{-2})

Let's denote the height of the first object as h_{1} and the height of the second object as h_{2}, where h_{2} = 4 x h_{1}

Given that the first object takes 2 seconds to reach the ground, we have:

Therefore, the time it will take for the second object to reach the ground is 4 seconds.

**2. Consider the following data for three planets:**

**Planet X:** orbital period = 2 years, semi-major axis = 3 AU**Planet Y:** orbital period = 5 years, semi-major axis = 5 AU**Planet Z:** orbital period = 10 years, semi-major axis = 8 AU

**According to Kepler's Third Law, which planet has the shortest orbital period?**

a) Planet X

b) Planet Y

c) Planet Z

d) Not enough information to determine

**Answer:** a) According to Kepler's Third Law, the square of the orbital period (T) is directly proportional to the cube of the semi-major axis (R) of the orbit. Mathematically:

T^{2} ∝ R^{3}

Let's calculate the ratios for each planet:

For Planet X:

(T_{X}^{2}) / (R_{X}^{3}) = (2^{2}) / (3^{3}) = 4 / 27

For Planet Y:

(T_{Y}^{2}) / (R_{Y}^{3}) = (5^{2}) / (5^{3}) = 25 / 125

For Planet Z:

(T_{Z}^{2}) / (R_{Z}^{3}) = (10^{2}) / (8^{3}) = 100 / 512

Comparing these ratios, we can see that Planet X indeed has the smallest value of (T_{X}^{2}) / (R_{X}^{3}), which indicates it has the shortest orbital period among the three planets.

**3. Consider two planets, X and Y, with the same mass but different radii. If planet X has twice the radius of planet Y, how does the acceleration due to gravity on planet X compare to that on planet Y?**

a) The acceleration on planet X is twice that on planet Y

b) The acceleration on planet X is half that on planet Y

c) The acceleration on planet X is one fourth that on planet Y.

d) The acceleration relationship depends on the planets' masses

**Answer:** c) The acceleration due to gravity is inversely proportional to the square of the distance from the centre of mass. This relationship is described by the formula:

g ∝ 1/r^{2}

Where

g is the acceleration due to gravity and

r is the distance from the centre of mass.

For planet Y, the acceleration due to gravity will be:

g_{Y} ∝ 1/r_{Y}^{2}

Given that planet X has twice the radius of planet Y (which means the distance r is doubled for planet X compared to planet Y), the acceleration due to gravity on planet X will be:

g_{X} ∝ 1/(2r_{Y})^{2}

g_{X} ∝ 1/4r_{Y}^{2}

g_{X}g_{Y} = 1/4

Comparing g_{X} and g_{Y}, we can see that g_{X} is one-fourth of g_{Y}, as indicated in option c.

**4. In the following question, you will find an assertion and a reason. Select the appropriate option that applies.**

**Assertion:** The acceleration due to gravity (g) at a depth below the Earth's surface is lower compared to the value of g at the surface.**Reason:** The mass of the Earth decreases as an object moves deeper below the surface, leading to a weaker gravitational force.

a) Both the assertion and reason are true, and the reason is the correct explanation of the assertion.

b) Both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

c) The assertion is true, but the reason is false.

d) The assertion is false, but the reason is true.

**Answer:** c) The assertion that the acceleration due to gravity (g) at a depth below the Earth's surface is lower compared to the value of g at the surface is true.

However, the reason provided – "The mass of the Earth decreases as an object moves deeper below the surface, leading to a weaker gravitational force" – is false. The mass of the Earth remains constant regardless of depth below the surface, and it's the distribution of mass that causes the decrease in gravity at greater depths. The true explanation is related to the distribution of mass, not a decrease in the Earth's overall mass.

**5. A spacecraft with a mass of 5000 kg is launched from Earth's surface into outer space. Calculate the weight of the spacecraft on Earth's surface and explain how this weight changes as the spacecraft moves away from Earth.**

a) Weight = 50,000 N; Weight decreases as the spacecraft moves away from Earth.

b) Weight = 50,000 N; Weight remains constant as the spacecraft moves away from Earth.

c) Weight = 5000 N; Weight decreases as the spacecraft moves away from Earth.

d) Weight = 5000 N; Weight remains constant as the spacecraft moves away from Earth.

**Answer:** a) The weight of an object on Earth's surface is the force with which it is attracted towards the centre of the Earth due to gravity. The formula to calculate weight (W) is given by:

W = mg

Where:

m is the mass of the object (5000 kg in this case),

g is the acceleration due to gravity on Earth's surface (9.8 ms^{-2}, or approximately 10 ms^{-2}).

Substitute these values into the formula:

W = 5000 kg × 10 ms^{-2} = 50,000 N

So, the weight of the spacecraft on Earth's surface is 40,000 N.

As the spacecraft moves away from Earth into outer space, the distance (r) between the spacecraft and the centre of the Earth increases. According to the universal law of gravitation, the force of gravity (F) between two objects is inversely proportional to the square of the distance (r) between their centres:

Since weight is the force of gravity on an object, as the spacecraft moves away from Earth, the force of gravity on it decreases. Therefore, the weight of the spacecraft decreases as it moves away from Earth.

**1. Who discovered the law of universal gravitation?**

Sir Isaac Newton discovered the law of universal gravitation in the seventeenth century. It asserts that every particle of matter in the cosmos attracts every other particle with a force equal to the product of their masses and inversely proportional to the square of the distance between their centers.

**2. How does gravity affect objects on Earth?**

Gravity attracts items to the centre of the Earth, giving them weight and keeping them fixed. When items are dropped, they tend to fall towards the ground.

**3. What factors affect the strength of gravity?**

The strength of gravity between two objects is determined by their masses and the distance between them. The gravitational pull is stronger when the objects have a higher mass and are closer together.

**4. How does the moon's gravity affect Earth?**

The moon's gravity pulls on Earth's seas, causing tides. This gravitational pull causes a bulge of water on the Earth's side facing the moon, as well as one on the opposing side.

**5. How does Newton's law of gravitation apply to celestial bodies?**

Newton's law of gravity describes how celestial things, such as planets, moons, and stars, are drawn to one another and kept in orbit. It contributes to scientists' understanding of the solar system and the universe.

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