Gravitation Solved Examples (Class 9 Physics)
Master gravitational force calculations using Newton's law of gravitation. These problems cover gravitational attraction between bodies, gravitational fiel
TL;DR: Master gravitational force calculations using Newton's law of gravitation. These problems cover gravitational attraction between bodies, gravitational…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
Master gravitational force calculations using Newton's law of gravitation. These problems cover gravitational attraction between bodies, gravitational fiel
Gravitation — Solved Numerical Examples (Step by Step)
Example 1: Calculate the gravitational force between Earth and the Sun. Given: Mass of Earth = 6 × 10^24 kg, Mass of Sun = 2 × 10^30 kg, Distance = 1.5 × 10^11 m, G = 6.67 × 10^-11 N·m²/kg².
Solution: Using Newton's law of gravitation: F = GMm/r²
F = (6.67 × 10^-11 × 2 × 10^30 × 6 × 10^24) / (1.5 × 10^11)²
Numerator = 6.67 × 2 × 6 × 10^-11 × 10^30 × 10^24 = 80.04 × 10^43 = 8.004 × 10^44 N·m²
Denominator = 2.25 × 10^22 m²
F = 8.004 × 10^44 / 2.25 × 10^22 = 3.558 × 10^22 N
Example 2: What is the gravitational force between two spheres of mass 2 kg each, separated by a distance of 0.5 m?
Solution: Using F = GMm/r²
G = 6.67 × 10^-11 N·m²/kg²
M = 2 kg, m = 2 kg, r = 0.5 m
F = (6.67 × 10^-11 × 2 × 2) / (0.5)²
F = (6.67 × 10^-11 × 4) / 0.25
F = 26.68 × 10^-11 / 0.25 = 106.72 × 10^-11 N = 1.0672 × 10^-9 N
Example 3: A person weighs 600 N on Earth. What would be their weight on the Moon if the gravitational field strength on the Moon is 1/6th that of Earth?
Solution: Weight on Earth = mg_Earth = 600 N
Gravitational field on Moon g_Moon = g_Earth / 6
Weight on Moon = m × g_Moon = m × (g_Earth / 6) = (m × g_Earth) / 6 = Weight_Earth / 6
Weight on Moon = 600 / 6 = 100 N
Example 4: Calculate the gravitational field strength at Earth's surface. Given: Mass of Earth = 6 × 10^24 kg, Radius of Earth = 6.4 × 10^6 m, G = 6.67 × 10^-11 N·m²/kg².
Solution: Gravitational field strength g = GM/r²
g = (6.67 × 10^-11 × 6 × 10^24) / (6.4 × 10^6)²
Numerator = 6.67 × 6 × 10^13 = 40.02 × 10^13 = 4.002 × 10^14 N·m²/kg
Denominator = 40.96 × 10^12 = 4.096 × 10^13 m²
g = 4.002 × 10^14 / 4.096 × 10^13 = 9.77 m/s²
Example 5: Two identical masses are separated by distance d. If the distance is reduced to d/2, how many times does the gravitational force increase?
Solution: Initial force F1 = GMm/d²
Final force F2 = GMm/(d/2)² = GMm/(d²/4) = 4GMm/d²
Ratio F2/F1 = (4GMm/d²) / (GMm/d²) = 4
Example 6: What is the mass of Earth if the gravitational field strength at its surface is 10 m/s², the radius is 6.4 × 10^6 m, and G = 6.67 × 10^-11 N·m²/kg²?
Solution: Using g = GM/R²
10 = (6.67 × 10^-11 × M) / (6.4 × 10^6)²
10 = (6.67 × 10^-11 × M) / (40.96 × 10^12)
M = 10 × 40.96 × 10^12 / 6.67 × 10^-11
M = 409.6 × 10^12 / 6.67 × 10^-11 = 61.4 × 10^23 = 6.14 × 10^24 kg
Tips
- Remember G = 6.67 × 10^-11 N·m²/kg² is a universal constant, not dependent on location.
- Gravitational force is always attractive and acts along the line joining the two masses.
- Weight = mg, where g varies with location; use g = 10 m/s² on Earth and g = 1.6 m/s² on the Moon for approximate calculations.
Frequently Asked Questions
Why do we use different values of g at different heights on Earth?
As you go higher, your distance r from Earth's center increases, so g = GM/r² decreases. At great heights, the effect becomes significant.
Is gravitational force between small objects (like two books) measurable?
Theoretically yes, but practically no because the force is extremely small (about 10^-9 N). Electromagnetic forces dominate at small scales.
More Physics Solved Examples
🤖 Stuck on any of these? Ask Syllab's free AI Tutor to explain step by step →