#astronomy

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 NCERT.

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

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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.

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PHYSICS

What is the minimum speed to escape from 3 R_E from Earth’s nter? ( g = 9.8 m/s², R_E = 6.4 × 10⁶m )

Given: What is the minimum speed to escape from 3 R_E from Earth’s nter? ( g = 9.8 m/s², R_E = 6.4 × 10⁶m ) Formula: v_e = √2 g R_E²/3 R_E = √2 × 9.8 × 6.4 × 10⁶/3. Substitution & Calculation: v_e = √4.181 × 10⁷approx 6.47 × 10³m/s approx 6.5 km/s . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

What is the escape speed from a planet of mass 4.8 × 10²⁴kg and radius 5 × 10⁶m ? ( G = 6.67 × 10⁻¹¹N m²/kg² )

Given: What is the escape speed from a planet of mass 4.8 × 10²⁴kg and radius 5 × 10⁶m ? ( G = 6.67 × 10⁻¹¹N m²/kg² ) These values define the system as per NCERT data. Formula: v_e = sqrt2 G M/R. This is standard NCERT relation. Substitution & Calculation: v_e = sqrtfrac2 × 6.67 × 10⁻¹¹ × 4.8 × 10²⁴⁵ × 10⁶. v_e = sqrt6.403 × 10⁷approx 8.0 × 10³m/s = 8.0 km/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A satellite orbits at 1.2 × 10⁷ m from a planet’s nter with period 3 hours. What is the planet’s mass? ( G = 6.67

Given: A satellite orbits at 1.2 × 10⁷ m from a planet’s nter with period 3 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 = 3 × 3600 = 10800 s, T² = 1.166 × 10⁸ s² . r³ = (1.2 × 10⁷)³ = 1.728 × 10²¹ m³ . M = frac4 × (3.14)² × 1.728 × 10²¹⁶.67 × 10⁻¹¹ × 1.166 × 10⁸. M = frac6.82 × 10²²⁷.78 × 10⁻³ approx 8.77 × 10²⁴ kg . 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 NCERT.

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

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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 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 NCERT.

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

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PHYSICS

Why can’t the gravitational force on a point mass outside a spherical shell be zero?

Outside a spherical shell, the gravitational force acts as if all the mass is concentrated at the nter. Since the mass is non-zero and the distance is finite, the force ( F = G M m/r² ) cannot be zero.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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Orbital Velocity and Time Period

Which of the following statements is correct about Kepler’s third law?

T2∝a3, meaning the period squared is proportional to the semi-major axis cubed, making option 2 correct. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Period squared is proportional to distance cubed. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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

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Kepler's Laws and Planetary Motion

What does Kepler’s third law imply about the relationship between orbital period and distance?

Kepler’s third law: the square of the orbital period (T2) is proportional to the cube of the semi-major axis (a3), i.e., T2∝a3. This relates time and distance for planetary motion.

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

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