#orbital speed

Universal Law of Gravitation

What ensures that a planet moves faster at perihelion than at aphelion?

Angular momentum conservation (mrv = constant) implies that at perihelion (smaller r), the speed v is greater than at aphelion (larger r), as derived from Kepler’s second law. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Conservation of angular momentum. 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

Universal Law of Gravitation

A satellite orbits Earth at 2.5RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 2.5RE, v = 9.8×6.4×1062.5. v = 2.509×107≈5.01×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.0 × 10³ m/s. 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

Gravitational Effects and Applications

A satellite orbits Earth at a height equal to RE/2. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

r = RE+RE2 = 1.5RE. v = GMEr = gRE21.5RE = gRE1.5. v = 9.8×6.4×1061.5 = 4.181×107. v≈6.47×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.5 × 10³ m/s. 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

Energy of Satellites and Binding Energy

A satellite orbits Earth at 19RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 19RE, v = 9.8×6.4×10619. v = 3.301×106≈1.82×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.8 × 10³ m/s. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

View Details

Energy of Satellites and Binding Energy

A satellite orbits Earth at 11RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 11RE, v = 9.8×6.4×10611. v = 5.698×106≈2.39×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.4 × 10³ m/s. 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 satellite orbits at 3RE from Earth’s center. What is its speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 3RE, v = 9.8×6.4×1063. v = 2.09×107≈4.57×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.6 × 10³ m/s. 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

Kepler's Laws and Planetary Motion

A satellite orbits Earth at 7RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 7RE, v = 9.8×6.4×1067. v = 8.966×106≈2.99×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.0 × 10³ m/s. 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

Work, Energy and Orbital Mechanics

A satellite orbits Earth at 17RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 17RE, v = 9.8×6.4×10617. v = 3.694×106≈1.92×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.9 × 10³ m/s. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

View Details

Gravitational Potential and Field

A satellite orbits Earth at 9RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 9RE, v = 9.8×6.4×1069. v = 6.964×106≈2.64×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6 × 10³ m/s. 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

Variation of g with Height and Depth

What happens to a satellite’s orbital speed if its altitude increases?

v = GMEr. As altitude increases, r (distance from Earth’s center) increases, reducing v since v∝1r. Thus, orbital speed decreases. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It decreases. 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

Variation of g with Height and Depth

A satellite orbits Earth at 15RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 15RE, v = 9.8×6.4×10615. v = 4.181×106≈2.04×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 × 10³ m/s. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

View Details

Universal Law of Gravitation

A satellite orbits Earth at 13RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 13RE, v = 9.8×6.4×10613. v = 4.826×106≈2.20×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.2 × 10³ m/s. 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