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#orbital speed

13 public questions tagged with this topic.

What is the frequency of revolution of an electron in the first orbit of a hydrogen atom if its speed is \( 2.2 \times 1

**Bohr's quantization** angular momentum L = m v r = n h/2π, for n=2 L=2h/2π= h/π=2.11×10⁻³⁴ J·s, for n=5 L=5h/2π, de Broglie λ = h/p, p= m v, for first orbit v=2.2×10⁶ m/s, λ= h/(m v)=6.6×10⁻³⁴/(9.1×10⁻³¹×2.2×10⁶)=3.3×10⁻¹⁰ m, circumference 2πr=3.33×10⁻¹⁰ m, one wavelength fits for n=1. v = (v/2π r) . v = (2.2 × 10⁶/2 × 3.14 × 5.3 × 10⁻¹¹) ≈ 6.6 × 10¹⁵ Hz . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 6.6 × 10¹⁵ Hz, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > De Broglie Hypothesis and Quantization in Bohr Model

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.