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Gravitation question

2019 · 10 Apr · Shift 2 · Q53
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Gravitation question

2019 · 10 Apr · Shift 2 · Q53

JEE MainPhysicsGravitationMCQ+4 / −1
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet? [Given ; Mass of planet = 8 × 1022 kg, Radius of planet = 2 × 106 m, Gravitational constant G = 6.67 × 10–11 Nm2 /kg2]
  1. A
    13
  2. B
    9
  3. C
    17
  4. D
    11
View written solutionFree

Correct answer: D

  1. Given data
  • Mass of planet: M=8×1022 kgM = 8\times 10^{22}\,\text{kg}M=8×1022kg
  • Radius of planet: R=2×106 mR = 2\times 10^6\,\text{m}R=2×106m
  • Height of spaceship above surface: h=20 km=2×104 mh = 20\,\text{km} = 2\times 10^4\,\text{m}h=20km=2×104m
  • Gravitational constant: G=6.67×10−11 N m2/kg2G = 6.67\times 10^{-11}\,\text{N m}^2/\text{kg}^2G=6.67×10−11N m2/kg2

So orbital radius is

r=R+h=2×106+2×104=2.02×106 mr = R+h = 2\times 10^6 + 2\times 10^4 = 2.02\times 10^6\,\text{m}r=R+h=2×106+2×104=2.02×106m
  1. Use formula for orbital period

For a circular orbit,

T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}T=2πGMr3​​

First compute GMGMGM:

GM=6.67×10−11×8×1022GM = 6.67\times 10^{-11}\times 8\times 10^{22}GM=6.67×10−11×8×1022 GM=53.36×1011=5.336×1012GM = 53.36\times 10^{11} = 5.336\times 10^{12}GM=53.36×1011=5.336×1012
  1. Compute r3r^3r3
r3=(2.02×106)3=2.023×1018r^3 = (2.02\times 10^6)^3 = 2.02^3\times 10^{18}r3=(2.02×106)3=2.023×1018 2.023≈8.24242.02^3 \approx 8.24242.023≈8.2424

Therefore,

r3≈8.2424×1018r^3 \approx 8.2424\times 10^{18}r3≈8.2424×1018
  1. Compute the quantity inside square root
r3GM=8.2424×10185.336×1012\frac{r^3}{GM} = \frac{8.2424\times 10^{18}}{5.336\times 10^{12}}GMr3​=5.336×10128.2424×1018​ ≈1.544×106\approx 1.544\times 10^6≈1.544×106

Thus,

r3GM≈1.544×106\sqrt{\frac{r^3}{GM}} \approx \sqrt{1.544\times 10^6}GMr3​​≈1.544×106​ ≈1242.6 s\approx 1242.6\,\text{s}≈1242.6s
  1. Find orbital period
T=2π×1242.6T = 2\pi \times 1242.6T=2π×1242.6 T≈7808 sT \approx 7808\,\text{s}T≈7808s

Convert to hours:

T=78083600≈2.17 hT = \frac{7808}{3600} \approx 2.17\,\text{h}T=36007808​≈2.17h
  1. Number of revolutions in 24 hours
N=242.17≈11.06N = \frac{24}{2.17} \approx 11.06N=2.1724​≈11.06

So the spaceship makes about 11 complete revolutions in 24 hours.

  1. Check options
  • A: 13 →\rightarrow→ incorrect
  • B: 9 →\rightarrow→ incorrect
  • C: 17 →\rightarrow→ incorrect
  • D: 11 →\rightarrow→ correct

Hence, the correct option is D.

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