Physics · Gravitation · chapter 7 of 14

Class 11 Physics Chapter 7: Gravitation case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
An Indian weather satellite of mass 1000 kg is placed in a circular orbit at a height of 600 km above the Earth's surface, so that the radius of its orbit is 7000 km. Engineers at the ground station use g = 9.8 m s⁻² at the Earth's surface and RE = 6400 km. The only force on the satellite is the Earth's gravity, which provides the centripetal force for its motion. The potential energy is taken as zero at infinity.
Part 1 · 1 mark
The orbital speed of the satellite is about
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Answer: (c) 7.57 km s⁻¹

V² = GME/r = gRE²/r = 9.8 × (6.4 × 10⁶)²/(7.0 × 10⁶), so V ≈ 7.57 × 10³ m s⁻¹. 8.28 km s⁻¹ wrongly uses √(gr) with the surface g; 10.7 km s⁻¹ is the escape speed from this orbit.

Part 2 · 1 mark
The period of the satellite is about
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Answer: (b) 97 min

T = 2πr/V = 2π × 7.0 × 10⁶/7573 ≈ 5.81 × 10³ s ≈ 97 min. 85 min is the period of an orbit just above the surface.

Part 3 · 2 marks
Find the total energy of the satellite in its orbit, and state what its negative sign tells us.
Answer and marking scheme

E = −GME m/2r = −mgRE²/2r = −1000 × 9.8 × (6.4 × 10⁶)²/(2 × 7.0 × 10⁶) ≈ −2.87 × 10¹⁰ J. The negative total energy means the satellite is bound to the Earth: it cannot reach infinity unless at least 2.87 × 10¹⁰ J is supplied.

  • E = −GM_E m/2r ≈ −2.87 × 10¹⁰ J (1)
  • negative: bound orbit, energy must be supplied to escape (1)
Or
Find the kinetic energy and the potential energy of the satellite in its orbit, and show how they are related.
Answer and marking scheme

K.E. = GME m/2r = mgRE²/2r ≈ 2.87 × 10¹⁰ J. P.E. = −GME m/r ≈ −5.73 × 10¹⁰ J. The potential energy is negative and twice the kinetic energy in magnitude: P.E. = −2 K.E.

  • K.E. ≈ 2.87 × 10¹⁰ J and P.E. ≈ −5.73 × 10¹⁰ J (1)
  • P.E. = −2 K.E. (1)
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