Skip to content
Login Register

#Earth's gravity

9 public questions tagged with this topic.

Escape Velocity and Bound Systems

What is the gravitational potential at a point 9.6×106m from Earth’s center due to Earth? ( ME\=6×1024kg,G\=6.67×10−11Nm

U = −GMEr. U = −6.67×10−11×6×10249.6×106. U = −4.002×10149.6×106 = −4.17×107J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -4.2 × 10⁷ J/kg. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

View Details
Orbital Velocity and Time Period

A 600kg satellite orbits Earth at 8RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−1

E = −GMEm2r. r = 8RE = 5.12×107m. E = −6.67×10−11×6×1024×6002×5.12×107. E = −2.401×10171.024×108≈−2.34×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.4 × 10⁹ J. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

View Details
Orbital Velocity and Time Period

A projectile is launched at 5km/s from Earth’s surface. What is its maximum distance from the center? (Escape speed = 11

12vi2−ve22 = −ve22REr. 12.5−62.72 = −62.72REr. rRE = 62.7250.22≈1.25. r = 1.25×6.4×106 = 8.0×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.0 × 10⁶ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

View Details
Variation of g with Height and Depth

A 3kg mass is moved from 4RE to 8RE from Earth’s center. What is the change in potential energy? (ME\=6×1024kg,RE\=6.4×1

ΔV = −GMEm(1r2−1r1). r1 = 2.56×107m, r2 = 5.12×107m. ΔV = −6.67×10−11×6×1024×3(15.12×107−12.56×107). ΔV = −1.201×1015(−1.953×10−8)≈2.34×107J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.4 × 10⁷ J. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

View Details
Universal Law of Gravitation

A body experiences g\=7.35m/s2 at a certain height above Earth. What is the height? (g0\=9.8m/s2,RE\=6.4×106m)

g(h) = g0(1+h/RE)2. 7.35 = 9.8(1+h/RE)2. (1+h/RE)2 = 9.87.35≈1.333. 1+h/RE = 1.333≈1.154. h/RE = 0.154. h = 0.154×6.4×106≈9.86×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.9 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

View Details