Find the value of acceleration due to gravity on the surface of the moon. Mass of the moon = 7 . 35 × 10 22 kg , Radius of the moon = 1 . 74 × 10 6 m , G = 6 . 67 × 10 - 11 Nm 2 kg - 2 .

The acceleration due to gravity on the moon can be calculated using: g = G M r 2 Where g is the acceleration due to gravity in ms - 2 , G is the gravitational constant = 6 . 67 × 10 - 11 Nm 2 kg - 2 , M is the mass of the moon = 7 . 35 × 10 22 kg , r is the radius of the moon in metres ( 1 . 74 × 10 6 ) . Hence, g = 6 . 67 × 10 - 11 × 7 . 35 × 10 22 ( 1 . 74 × 10 6 ) × 1 . 74 × 10 6 g = 1 . 62 m s - 2 . The value of g depends on the mass of the massive body and its radius and its value varies from one body to another. The value of g on moon is constant.

A body is dropped from the top of a tower. During the last second of its fall, it covers 16 25 t h of the height of the tower. Calculate the height of the tower.

Initial velocity, u = 0 Let the height of the tower be ‘ h ’. Suppose it takes ‘ n ’ second to reach the ground. Thus, in n t h second the body covers a height of 16 25 h . In ( n - 1 ) t h second, the body covers a height of 1 - 16 25 h = 25 - 16 25 h = 9 25 h It starts from rest, thus, We know that the equation to find the distance travelled in ( n - 1 ) t h second is S n = u + a 2 ( n – 1 ) 2 9 25 h = 0 + 9 . 8 2 ( n – 1 ) 2 9 25 h = 4 . 9 ( n – 1 ) 2 → 1 Again, the body reaches the ground in ‘ n ’ second. So, the distance travelled in ‘ n ’ second is, h = 0 + g 2 n 2 h = 9 . 8 2 n 2 h = 4 . 9 n 2 → 2 Substituting value of h in Eq 1 we get, 9 25 × 4 . 9 n 2 = 4 . 9 ( n – 1 ) 2 9 × 4 . 9 n 2 25 = 4 . 9 ( n – 1 ) 2 9 25 n 2 = ( n – 1 ) 2 9 n 2 = 25 ( n – 1 ) 2 9 n 2 = 25 n 2 – 2 n + 1 = 25 n 2 - 50 n + 25 25 n 2 - 50 n + 25 - 9 n 2 = 0 16 n 2 - 50 n + 25 = 0 Use the quadratic formula n = - b ± b 2 - 4 a c 2 a a = 16 b = - 50 c = 25 n = - - 50 ± - 50 2 - 4 × 16 × 25 2 × 16 n = 50 ± 30 32 n = 50 + 30 32 = 5 2 = 2 . 5 n = 50 - 30 32 = 5 8 = 0 . 625 Here, n = 0 . 625 s is not possible as it is greater than 1 . So, the time in which the body hits the ground is 2 . 5 s . Putting value of n in equation ( 2 ) , we get h = 4 . 9 n 2 = 4 . 9 × 2 . 5 × 2 . 5 = 30 . 625 m . Hence, height of the tower is equal to 30 . 625 m .

The value of ' G ' on the surface of earth is 6 . 67 × 10 - 11 N m ² / kg ² . What will be its value on the surface of moon?

The value of ' G ' on the surface of the moon is 6 . 67 × 10 - 11 N m ² / kg ² . G is Universal gravitational constant, it is independent of the nature of the particle, the medium between the particles, and time. Its value is constant anywhere in the Universe, and hence it's called 'Universal'. So the value of G is the same everywhere.

E M B I B E

Physics>Chapter 3 - Gravitation>EXAM PRACTICE QUESTIONS>Q 24

MEDIUM

9th CBSE

IMPORTANT

Earn 100

An apple falls towards the earth due to the gravitational force of attraction. The apple also attracts the earth with the same force. Why do we not see the earth rising towards the apple? Explain.

Important Questions on Gravitation

MEDIUM

9th CBSE

IMPORTANT

Physics>Chapter 3 - Gravitation>EXAM PRACTICE QUESTIONS>Q 25

Describe the state of weightlessness with the help of a suitable example.

MEDIUM

9th CBSE

IMPORTANT

Physics>Chapter 3 - Gravitation>EXAM PRACTICE QUESTIONS>Q 26

(a) Show that the value of 'g' decreases as we go to higher altitudes above the surface of the earth. (b) Mention one practical application of this property of 'g'.

MEDIUM

9th CBSE

IMPORTANT

Physics>Chapter 3 - Gravitation>EXAM PRACTICE QUESTIONS>Q 27

Find the value of acceleration due to gravity on the surface of the moon. Mass of the moon = $7.35 \times 10^{22} kg$ , Radius of the moon = $1.74 \times 10^{6} m$ , G = $6.67 \times 10^{- 11} {Nm}^{2} {kg}^{- 2}$ .

HARD

9th CBSE

IMPORTANT

Physics>Chapter 3 - Gravitation>EXAM PRACTICE QUESTIONS>Q 28

A body is dropped from the top of a tower. During the last second of its fall, it covers

{(\frac{16}{25})}^{t h}

of the height of the tower. Calculate the height of the tower.

HARD

9th CBSE

IMPORTANT

Physics>Chapter 3 - Gravitation>EXAM PRACTICE QUESTIONS>Q 29

An object is dropped from an aircraft when it is directly above a target at a height of

1000 m

. The aircraft is moving with a velocity of

500 km {hr}^{- 1}

. Will the bomb hit the target, if not, by how much distance it will miss the target?

HARD

9th CBSE

IMPORTANT

Physics>Chapter 3 - Gravitation>EXAM PRACTICE QUESTIONS>Q 30

A rocket is launched to travel vertically upwards with a constant velocity of $20 m s^{- 1}$ . After travelling for $35 seconds$ the rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body. What is the total height achieved by it? After what time of its launch will it come back to the earth?