Gravitation - Physics For Class IX (Science Group) - Numericals

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Physics For Class IX (Science Group)
UNIT 6: GRAVITATION
Numericals

WORKED EXAMPLES

Worked Example 2

2. Calculate the weight of Rumaisa, who has a mass of 65 kg standing at the ground. The strength of gravitational field on Rumaisa is 10 Newton per kilogram?
Solution:
Step 1: Write down known quantities and quantities to be found.

• m = 65 kg
• g = 10 Nkg^-1
• W = ?

Step 2: Write down formula and rearrange if necessary

W = mg

Step 3: Put the values in formula and calculate

W = 65 kg x 10 Nkg^-1

Ans: Hence, the weight of Rumaisa is 650 Newton.

Worked Example 3

3. Calculate the acceleration due to gravity on a planet that has mass two times to the mass of Earth and radius 1.5 times to the radius of Earth. If the acceleration due to gravity on the surface of Earth is 10 ms^-2. Calculate acceleration due to gravity on the planet?
Solution
Step 1: Write down known quantities and quantities to be found.

• Mass of the planet = M_p = 2 M_E
• Mass of the Earth = M_E = 6.0 x 10²⁴ kg
• Radius of the planet = R_p = 1.5 R _E
• Radius of the Earth = R _E = 6.38 x 10⁶ m
• Universal gravitational constant = G = 6.673 x 10^-11 Nm² kg^-2
• Acceleration due to gravity on the Earth = g _E =10 ms^-2
• Acceleration due to gravity on the planet = g_p = ?

Step 2: Write down formula and rearrange if necessary
For the Earth:

Ans: Hence, acceleration due to gravity on the planet is 8.74 ms^-2.

TEXT BOOK EXERCISE

Section (B) Structured Questions
Newton’s Law of Gravitation
4. Determine the gravitational force of attraction between Urwa and Ayesha standing at a distance of 50 m apart. The mass of Urwa is 60 kg and that of Ayesha is 70 kg.
Solution:
Data:

1. Mass of Urwa = m₁ = 60 kg
2. Mass of Ayesha = m₂ = 70 kg
3. Distance between Urwa and Ayesha = r = 50 m
4. Universal gravitation = G = 6.673 x 10^-11 Nm²kg^-2
5. Gravitational force = F = ?

Ans: Hence the gravitational force of attraction between Urwa and Ayesha is 1.121064 x 10^-10 N.

Weight
6.b) Weight of Rani is 450 N at the surface of Earth. Find her mass?
Solution:
Data:

• Weight of Rani = 450 N
• Acceleration due to gravity = g = 10 ms^-2
• Mass of Rani = ?

Working Formula:

W = mg

Calculation:

450 = m (10)

m = ⁴⁵⁰/₁₀ = 45 kg

Ans: Hence her mass is 45 kg.

7. Weight of Naveera is 700 N on the Earth's surface. What will be Naveera's weight at the surface of Moon?
Solution:
Data:

• Weight of Naveera on the Earth's surface = W_E = 700 N
• Acceleration due to gravity on the surface of Earth = g_E = 10 Nkg^-1
• Acceleration due to gravity on the surface of Moon = g_M = 1.6 Nkg^-1
• Naveera's weight at the surface of Moon = W_M = ?

Working Formula:

W = mg

Calculation:
To calculate Naveera's weight at the surface of Moon, We need to find her mass. To find her mass we apply the formula:

W_E = mg_E

700 = m (10)

m = ⁷⁰⁰ / ₁₀ = 70 kg

Now, to find her weight at the surface of Moon,

W_M = mg_M

W_M = 70 x 1.6 = 112 N

Ans: Hence Naveera's weight at the surface of Moon will be 112 N.

Mass of Earth
10. If “M ” is the mass of Earth, “R ” radius of Earth, “G” is universal gravitational constant, then find acceleration due to gravity “g”;
i) On the surface of Earth.
ii) At the centre of Earth.
Solution:
i) On the surface of Earth:
We consider a body of mass 'm' on the earth's surface. The gravitational force of attraction between the body and the earth are equal to the weight of the body.

ii) At the centre of Earth:
The force due to any portion of the Earth at the centre will be cancelled due to the portion opposite to it. Thus, the gravitational force at the centre on any body will be 0. Since, from Newton's law, we know F = mg. Since the mass m of an object can never be 0. Therefore, when F = 0, then g has to be 0. Thus the value of g is zero at the centre of the Earth.

11. A planet has mass four times of Earth and radius two times that of Earth. If the value of “g” on the surface of Earth is 10 ms^–2. Calculate acceleration due to gravity on the planet.
Solution:
Data:

• Mass of the planet = M_p = 4M_e
• Radius of the planet = r_p = 2R_e
• Acceleration due to gravity on the surface of Earth = g = 10 ms^–2
• Acceleration due to gravity on the surface of Earth = g_p = ?

Artificial Satellite
13.a) Calculate the speed of a satellite which orbits the Earth at an altitude of 400 kilometers above Earth's surface.
Solution:
Data:

• Mass = M_Earth = M = 6.0 x 10²⁴ kg
• Radius = R_Earth = R = 6.38 x 10⁶ m
• Altitude = h = 400 km = 400 x 10³ m = 4 x 10⁵ m
• Universal gravitation = 6.673 x 10^-11 Nm²kg^-2
• v =?

Ans: Hence, the orbital speed of satellite is 5.905 x 10⁷ ms^-1

Self Assessment Questions

Q.5: The strength of gravity on the Moon is 1.6 N kg^-1 . If an astronaut’s mass is 80 kg on Earth, what would it be (weight) on the Moon?
Solution:
Data:

• g = 1.6 N kg^-1
• m = 80 kg
• W = ?

Working Formula:

W = mg

Calculation:

W = 80 x 1.6 = 128 N

Ans: Hence, The weight of the astronaut on the moon is 128 N.

Q.7: What will be the value of acceleration due to gravity on the surface of earth if its radius reduces to half?
Solution:

Q.8: What will be acceleration due to gravity on the surface of earth f its mass reduces by 25%?
Solution:

Q.9: What will be the mass of a planet whose radius is 20% of the radius of earth?
Solution:

Activity (Text book Pg No. 143)

1. Calculate the mass of the Earth if acceleration due to gravity; g = 9.8 ms^–2.
Solution:
Data:

• g = 9.8 ms^–2.
• R_E = 6.38 x 10⁶m
• G = 6.673 x 10^-11Nm²kg^-2
• M_E 

M_E = 6.0 x 10^{24 kg}

Ans: Hence, the mass of Earth is 6.0 x 10²⁴ kg.