JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Gravitation

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A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the plant acts on the spaceship. What will be the number of complete revolutions made by the spaceship in 24 hours around the plane?
[Given: Mass of plane

= 8 \times 1 0^{22} kg

, Radius of planet

= 2 \times 1 0^{6} m

, Gravitational constant

G = 6.67 \times 1 0^{- 11} Mn^{2} / kg^{2}

]

(A)9
(B)11✓ Correct
(C)13
(D)17

Answer: (B)

Solution

The orbit radius is

r = R + h = 2 \times 1 0^{6} + 20 \times 1 0^{3} = 2.02 \times 1 0^{6} m

. The period follows from

T = 2 π \frac{r ^{3}}{GM} = 2 π \frac{( 2.02 \times 1 0 ^{6} ) ^{3}}{( 6.67 \times 1 0 ^{- 11} ) ( 8 \times 1 0 ^{22} )} \approx 7.8 \times 1 0^{3} s

. In

24

hours the craft completes

\frac{24 \times 3600}{7800} \approx 11

full revolutions.

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Gravitation

Solution

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