Skip to content
Login Register

#orbital mechanics

34 public questions tagged with this topic.

PHYSICS

A satellite near Earth has a period of 82 minutes. What is its period at h = 9 R_E ? ( R_E = 6.4 × 10⁶ m )

Given: A satellite near Earth has a period of 82 minutes. What is its period at h = 9 R_E ? ( R_E = 6.4 × 10⁶ m ) These values define the system as per NCERT data. Formula: T_0² = k R_E³, h = 9 R_E, r = 10 R_E. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: T² propto (R_E + h)³ . . T² = k (10 R_E)³ = 1000 k R_E³ . T = T_0 sqrt1000 = 82 × 31.62 approx 2593 min . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per N

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

View Details
PHYSICS

A planet orbits the Sun with a period of 7 years. If Earth’s orbital radius is 1.5 × 10¹¹m, what is its semi-major axis?

Given: A planet orbits the Sun with a period of 7 years. If Earth’s orbital radius is 1.5 × 10¹¹m, what is its semi-major axis? These values define the system as per NCERT data. Formula: Kepler’s third law: T_p²/T_E² = a_p³/a_E³. This is standard NCERT relation. Substitution & Calculation: T_E = 1 year, T_p = 7 years, a_E = 1.5 × 10¹¹m . 7²/1² = fraca_p³(1.5 × 10¹¹)³ . 49 = fraca_p³³.375 × 10³³. a_p³ = 49 × 3.375 × 10³³= 1.65375 × 10³⁵. a_p = (1.65375 × 10³⁵)^{1/3 approx 5.49 × 10¹¹m . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Gravitation, Topic: Kepler's third law T² ∝ a³, orbital period, semi-major axis and planetary motion. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

View Details
PHYSICS

How much energy is required to move a 100 kg satellite from 4 R_E to 8 R_E from Earth’s nter? ( M_E = 6 × 10²⁴ kg,

Given: How much energy is required to move a 100 kg satellite from 4 R_E to 8 R_E from Earth’s nter? ( M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg² ) These values define the system as per NCERT data. Formula: Δ E = -G M_E m (1/r_2 - 1/r_1). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r_1 = 2.56 × 10⁷ m, r_2 = 5.12 × 10⁷ m . Δ E = -6.67 × 10⁻¹¹ × 6 × 10²⁴ × 100 (1/5.12 × 10⁷- 1/2.56 × 10⁷) . Δ E = -4.002 × 10¹⁶(-1.953 × 10⁻⁸) approx 7.82 × 10⁸ J . Result: The computed

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

View Details
PHYSICS

A satellite orbits a planet at 1.5 × 10⁷ m from its nter with a period of 4 hours. What is the planet’s mass? ( G =

Given: A satellite orbits a planet at 1.5 × 10⁷ m from its nter with a period of 4 hours. What is the planet’s mass? ( G = 6.67 × 10⁻¹¹ N m²/kg² ) These values define the system as per NCERT data. Formula: M = 4π² r³/G T². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: T = 4 × 3600 = 14400 s, T² = 2.0736 × 10⁸ s² . r³ = (1.5 × 10⁷)³ = 3.375 × 10²¹ m³ . M = frac4 × (3.14)² × 3.375 × 10²¹⁶.67 × 10⁻¹¹ × 2.0736 × 10⁸. M = frac1.331 × 10²³¹.383 × 10⁻² approx 9.63 × 10²⁴ kg . Result: The

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

View Details
Gravitational Effects and Applications

How much energy is required to move a 700kg satellite from 11RE to 22RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,

ΔE = −GMEm(1r2−1r1). r1 = 7.04×107m, r2 = 1.408×108m. ΔE = −6.67×10−11×6×1024×700(11.408×108−17.04×107). ΔE = −2.801×1017(−7.102×10−9)≈1.99×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 × 10⁹ J. 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

How much energy is required to move a 500kg satellite from 5RE to 10RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G

ΔE = −GMEm(1r2−1r1). r1 = 3.2×107m, r2 = 6.4×107m. ΔE = −6.67×10−11×6×1024×500(16.4×107−13.2×107). ΔE = −2.001×1017(−1.5625×10−8)≈3.13×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.1 × 10⁹ J. 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 projectile is launched from Earth with a speed of 15km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

Total energy: E = 12mvi2−GMEmRE. At infinity: E = 12mvf2, ve = 2GMERE. 12vi2−12ve2 = 12vf2. vf2 = (15)2−(11.2)2 = 225−125.44 = 99.56. vf = 99.56≈10km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 km/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 projectile is launched at 1km/s from Earth’s surface. What is its maximum distance from the center? (Escape speed = 11

12vi2−ve22 = −ve22REr. 0.5−62.72 = −62.72REr. rRE = 62.7262.22≈1.008. r = 1.008×6.4×106≈6.45×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.5 × 10⁶ m. 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