Q1: Determine the gravitational force of attraction between Urwa and Ayesha standing at a distance of 50m apart. The mass of Urwa is 60kg and that of Ayesha is 70kg.
Given:
Mass of Urwa, m₁ = 60 kg
Mass of Ayesha, m₂ = 70 kg
Distance, r = 50 m
G = 6.67 × 10⁻¹¹ Nm²/kg²
Formula: F = G × (m₁ × m₂) / r²
Calculation:
F = (6.67 × 10⁻¹¹ × 60 × 70) / (50)²
F = (6.67 × 10⁻¹¹ × 4200) / 2500
F = (2.8014 × 10⁻⁷) / 2500
F = 1.12 × 10⁻¹⁰ N
Answer: 1.12 × 10⁻¹⁰ N
Mass of Urwa, m₁ = 60 kg
Mass of Ayesha, m₂ = 70 kg
Distance, r = 50 m
G = 6.67 × 10⁻¹¹ Nm²/kg²
Formula: F = G × (m₁ × m₂) / r²
Calculation:
F = (6.67 × 10⁻¹¹ × 60 × 70) / (50)²
F = (6.67 × 10⁻¹¹ × 4200) / 2500
F = (2.8014 × 10⁻⁷) / 2500
F = 1.12 × 10⁻¹⁰ N
Answer: 1.12 × 10⁻¹⁰ N
Q2: Weight of Naveera is 700N on the Earth's surface. What will be Naveera's weight at the surface of Moon?
Given:
Weight on Earth, Wₑ = 700 N
gₘ (Moon) = 1/6 × gₑ (Earth)
Formula: Weight on Moon = (1/6) × Weight on Earth
Calculation:
Wₘ = 700 × (1/6)
Wₘ = 116.67 N
Answer: 116.67 N
Weight on Earth, Wₑ = 700 N
gₘ (Moon) = 1/6 × gₑ (Earth)
Formula: Weight on Moon = (1/6) × Weight on Earth
Calculation:
Wₘ = 700 × (1/6)
Wₘ = 116.67 N
Answer: 116.67 N
Q3: Weight of Rani is 450N at the surface of Earth. Find her mass?
Given:
Weight, W = 450 N
g = 10 m/s² (standard value)
Formula: W = m × g
m = W / g
Calculation:
m = 450 / 10
m = 45 kg
Answer: 45 kg
Weight, W = 450 N
g = 10 m/s² (standard value)
Formula: W = m × g
m = W / g
Calculation:
m = 450 / 10
m = 45 kg
Answer: 45 kg
Q4: 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 10ms⁻². Calculate acceleration due to gravity on the planet.
Given:
Mass of planet, Mₚ = 4Mₑ
Radius of planet, Rₚ = 2Rₑ
gₑ = 10 m/s²
Formula: g = GM/R²
gₚ/gₑ = (Mₚ/Mₑ) × (Rₑ/Rₚ)²
Calculation:
gₚ/gₑ = 4 × (1/2)²
gₚ/gₑ = 4 × 1/4
gₚ/gₑ = 1
gₚ = gₑ = 10 m/s²
Answer: 10 m/s²
Mass of planet, Mₚ = 4Mₑ
Radius of planet, Rₚ = 2Rₑ
gₑ = 10 m/s²
Formula: g = GM/R²
gₚ/gₑ = (Mₚ/Mₑ) × (Rₑ/Rₚ)²
Calculation:
gₚ/gₑ = 4 × (1/2)²
gₚ/gₑ = 4 × 1/4
gₚ/gₑ = 1
gₚ = gₑ = 10 m/s²
Answer: 10 m/s²
Q5: Evaluate the acceleration due to gravity in terms of mass of Earth "Mₑ", radius of Earth "Rₑ" and universal gravitational constant "G":
i) At a distance, twice the Earth's radius.
ii) At a distance, one half the Earth's radius.
Given:
Mass of Earth = Mₑ
Radius of Earth = Rₑ
G = Universal gravitational constant
Formula: g = GM/r²
i) At distance twice Earth's radius:
r = 2Rₑ
g₁ = G × Mₑ / (2Rₑ)²
g₁ = G × Mₑ / 4Rₑ²
g₁ = (1/4) × (GMₑ/Rₑ²)
g₁ = g/4 (where g = GMₑ/Rₑ²)
ii) At distance one half Earth's radius:
r = Rₑ/2
g₂ = G × Mₑ / (Rₑ/2)²
g₂ = G × Mₑ / (Rₑ²/4)
g₂ = 4 × (GMₑ/Rₑ²)
g₂ = 4g
Answer:
i) g/4
ii) 4g
Mass of Earth = Mₑ
Radius of Earth = Rₑ
G = Universal gravitational constant
Formula: g = GM/r²
i) At distance twice Earth's radius:
r = 2Rₑ
g₁ = G × Mₑ / (2Rₑ)²
g₁ = G × Mₑ / 4Rₑ²
g₁ = (1/4) × (GMₑ/Rₑ²)
g₁ = g/4 (where g = GMₑ/Rₑ²)
ii) At distance one half Earth's radius:
r = Rₑ/2
g₂ = G × Mₑ / (Rₑ/2)²
g₂ = G × Mₑ / (Rₑ²/4)
g₂ = 4 × (GMₑ/Rₑ²)
g₂ = 4g
Answer:
i) g/4
ii) 4g
Q6: Calculate the speed of a satellite which orbits the Earth at an altitude of 400 kilometers above Earth's surface.
Given:
Altitude, h = 400 km = 400,000 m
Radius of Earth, Rₑ = 6.37 × 10⁶ m
Mass of Earth, Mₑ = 5.98 × 10²⁴ kg
G = 6.67 × 10⁻¹¹ Nm²/kg²
Formula: v = √(GMₑ/(Rₑ + h))
Calculation:
r = Rₑ + h = 6.37 × 10⁶ + 0.4 × 10⁶
r = 6.77 × 10⁶ m
v = √[(6.67 × 10⁻¹¹ × 5.98 × 10²⁴) / (6.77 × 10⁶)]
v = √[(3.988 × 10¹⁴) / (6.77 × 10⁶)]
v = √(5.89 × 10⁷)
v = 7.67 × 10³ m/s
v = 7.67 km/s
Answer: 7.67 km/s or 7670 m/s
Altitude, h = 400 km = 400,000 m
Radius of Earth, Rₑ = 6.37 × 10⁶ m
Mass of Earth, Mₑ = 5.98 × 10²⁴ kg
G = 6.67 × 10⁻¹¹ Nm²/kg²
Formula: v = √(GMₑ/(Rₑ + h))
Calculation:
r = Rₑ + h = 6.37 × 10⁶ + 0.4 × 10⁶
r = 6.77 × 10⁶ m
v = √[(6.67 × 10⁻¹¹ × 5.98 × 10²⁴) / (6.77 × 10⁶)]
v = √[(3.988 × 10¹⁴) / (6.77 × 10⁶)]
v = √(5.89 × 10⁷)
v = 7.67 × 10³ m/s
v = 7.67 km/s
Answer: 7.67 km/s or 7670 m/s
