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Class 11 CBSE | Gravitation Numericals (Answers Visible)

Gear Institute

Class 11 CBSE

Gravitation

Numericals on Newton’s Law of Gravitation & Variation of g (with Height & Depth)

Exact questions and solutions as requested — answers are visible (not hidden).

⚙️ NUMERICALS ON NEWTON’S LAW OF GRAVITATION

1) Basic

Find the gravitational force between two masses of 10 kg and 20 kg separated by a distance of 5 m. (G = 6.67 × 10⁻¹¹ N·m²/kg²)

F = G m₁ m₂ / r² = (6.67 × 10⁻¹¹ × 10 × 20) / 5² = 5.34 × 10⁻¹⁰ N.

2) Basic

Two spheres of equal mass m = 5 kg attract each other with a force of 2.67 × 10⁻⁹ N. Find the distance between them.

F = G m² / r² ⇒ r = √(G m² / F) = √((6.67 × 10⁻¹¹ × 25) / (2.67 × 10⁻⁹)) = 0.79 m.

3) Basic

A body weighs 60 N on the surface of the Earth. Find its mass. (g = 9.8 m/s²)

m = W / g = 60 / 9.8 = 6.12 kg.

4) Intermediate

Calculate the gravitational force between Earth and a man of mass 70 kg. (M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10⁶ m)

F = G M_E m / R_E² = (6.67 × 10⁻¹¹ × 6 × 10²⁴ × 70) / (6.4 × 10⁶)² = 686 N.

5) Intermediate

At what distance from the Earth’s center will the gravitational force be 1/4th of its value at the surface?

Since F ∝ 1/r², F′/F = (R/r)² = 1/4 ⇒ r = 2R (twice Earth’s radius).

6) Intermediate

A satellite revolves around the Earth at a height where g′ = g/4. Find the height h above the surface of Earth. (g′ = g (R/(R+h))²)

1/4 = (R/(R+h))² ⇒ R+h = 2R ⇒ h = R = 6.4 × 10⁶ m.

🌍 VARIATION OF ACCELERATION DUE TO GRAVITY

With Height (Above Surface)

Find the value of g at a height h = 640 km above Earth’s surface. (R = 6400 km)

g′ = g (R/(R+h))² = 9.8 (6400/7040)² = 7.94 m/s².

If the value of g decreases by 1%, find the height above the Earth’s surface.

g′/g = 0.99 = (R/(R+h))² ⇒ (R+h)/R = √(1/0.99) = 1.005 ⇒ h = 0.005R = 0.005 × 6400 = 32 km.

With Depth (Below Surface)

Find the value of g at a depth of 1600 km. (R = 6400 km)

g′ = g (1 − d/R) = 9.8 (1 − 1600/6400) = 9.8 × 0.75 = 7.35 m/s².

At what depth below the surface will the value of g become half of its surface value?

g′/g = 1/2 = 1 − d/R ⇒ d = R/2 = 3200 km.

💡 ADVANCED / CONCEPTUAL

11)

A body weighs 100 N on Earth’s surface. Find its weight at a height equal to the Earth’s radius.

g′ = g (R/(R+h))² = g (R/2R)² = g/4 ⇒ W′ = 100/4 = 25 N.

12)

The acceleration due to gravity on the surface of planet X is 5 m/s², and its radius is half that of Earth. Find the density ratio of planet X to Earth. (g = (4/3)π G R ρ)

(g_X/g_E) = (R_X ρ_X)/(R_E ρ_E) ⇒ 5/9.8 = (1/2)(ρ_X/ρ_E) ⇒ ρ_X/ρ_E = 1.02 (approximately).

13)

A rocket goes to a height where the value of g becomes 6.13 m/s². Find the height h if R_E = 6400 km.

(g′/g) = (R/(R+h))² ⇒ √(6.13/9.8) = R/(R+h) ⇒ R+h = 1.26R ⇒ h = 0.26R = 1664 km.

14)

Find the effective value of g at a place on the latitude 45°. (g′ = g − R ω² cos²θ, where ω = 7.27 × 10⁻⁵ rad/s, R = 6.4 × 10⁶ m)

R ω² ≈ 0.034 m/s²; g′ = 9.8 − 0.034 cos²45° = 9.8 − 0.017 = 9.783 m/s².

15)

At what height will the value of g become 3/4th of its surface value?

(g′/g) = (R/(R+h))² = 3/4 ⇒ (R+h)/R = √(4/3) = 1.155 ⇒ h = 0.155R = ≈ 992 km.

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