Gravitation – IIT/JEE Solved Problems

Phase 1 • Page 2 • Height, Depth & Satellite Basics


Problem 6: g at Depth d

Q: Find acceleration due to gravity at a depth equal to one-third the radius of Earth.

Solution:

Inside Earth:
gd = g (1 − d/R)

Here, d = R/3
gd = g (1 − 1/3) = 2g/3

Key Idea: Inside Earth, g decreases linearly with depth.


Problem 7: Height Where Weight Becomes Half

Q: At what height above Earth’s surface does a body’s weight become half of its value on the surface?

Solution:

Weight ∝ g, so:
gh = g/2

gh = g (R² / (R + h)²)

R² / (R + h)² = 1/2
(R + h) = √2 R
h = (√2 − 1)R

IIT Tip: Do not approximate √2 unless options allow.


Problem 8: Orbital Speed of Satellite

Q: Find orbital speed of a satellite very close to Earth’s surface.

Solution:

v = √(GM/R)

Since g = GM/R²,
v = √(gR)

Shortcut: Near Earth’s surface, use √(gR) directly.


Problem 9: Time Period of Satellite

Q: Find the time period of a satellite orbiting close to Earth’s surface.

Solution:

T = 2π √(R³/GM)

Using g = GM/R²,
T = 2π √(R/g)

Remember: This formula is frequently asked directly.


Problem 10: Satellite at Height R

Q: A satellite is orbiting Earth at a height equal to Earth’s radius. Find the ratio of its orbital speed to that near Earth’s surface.

Solution:

At surface: v₁ = √(GM/R)
At height R: distance = 2R
v₂ = √(GM/2R)

v₂ / v₁ = √(1/2)

Trap Alert: Always write distance from centre, not height.


Phase-1 Skill Upgrade

  • Convert height/depth to distance from centre
  • Use g–GM–R relations smartly
  • Avoid plugging values early
  • Check proportionality first

Phase 1 • Page 2 Completed Successfully

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