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MCQEasyJEE 2024Variation with Altitude & Depth

JEE Physics 2024 Question with Solution

The gravitational potential at a point above the surface of Earth is −5.12×107 J/kg-5.12 \times 10^7 \, \text{J/kg} and the acceleration due to gravity at that point is 6.4 m/s26.4 \, \text{m/s}^2. Assume that the mean radius of Earth to be 6400 km6400 \, \text{km}. The height of this point above the Earth’s surface is:

  • A

    1600 km1600 \, \text{km}

  • B

    540 km540 \, \text{km}

  • C

    1200 km1200 \, \text{km}

  • D

    1000 km1000 \, \text{km}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Gravitational potential at the point is V=−5.12×107 J/kgV = -5.12 \times 10^7 \, \text{J/kg}, acceleration due to gravity there is g′=6.4 m/s2g' = 6.4 \, \text{m/s}^2, and Earth’s radius is R=6400 km=6400×103 mR = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m}.

Find: The height hh above Earth’s surface.

Use the relations:

V=−GMR+hV = -\frac{GM}{R+h}

and

g′=GM(R+h)2g' = \frac{GM}{(R+h)^2}

From these two equations, divide the magnitude of potential by gravitational acceleration:

GM/(R+h)GM/(R+h)2=5.12×1076.4\frac{GM/(R+h)}{GM/(R+h)^2} = \frac{5.12 \times 10^7}{6.4}

So,

R+h=5.12×1076.4=8×106 mR+h = \frac{5.12 \times 10^7}{6.4} = 8 \times 10^6 \, \text{m}

Now substitute R=6400×103 m=6.4×106 mR = 6400 \times 10^3 \, \text{m} = 6.4 \times 10^6 \, \text{m}:

h=8×106−6.4×106h = 8 \times 10^6 - 6.4 \times 10^6 h=1.6×106 mh = 1.6 \times 10^6 \, \text{m}

Converting to kilometres,

h=1600 kmh = 1600 \, \text{km}

Therefore, the height above Earth’s surface is 1600 km1600 \, \text{km}. The correct option is A. The solution states option C, but its working clearly gives 1600 km1600 \, \text{km}, which matches option A.

Direct Ratio Trick

Given: V=−5.12×107 J/kgV = -5.12 \times 10^7 \, \text{J/kg}, g′=6.4 m/s2g' = 6.4 \, \text{m/s}^2, R=6400 kmR = 6400 \, \text{km}.

Find: Height hh.

Since

V=−GMrV = -\frac{GM}{r}

and

g′=GMr2g' = \frac{GM}{r^2}

we get directly:

∣V∣g′=r\frac{|V|}{g'} = r

where r=R+hr = R+h. This works because GMGM cancels immediately.

Now calculate:

r=5.12×1076.4=8×106 mr = \frac{5.12 \times 10^7}{6.4} = 8 \times 10^6 \, \text{m}

Then,

h=r−R=8×106−6.4×106=1.6×106 m=1600 kmh = r - R = 8 \times 10^6 - 6.4 \times 10^6 = 1.6 \times 10^6 \, \text{m} = 1600 \, \text{km}

Therefore, the correct option is A.

Common mistakes

  • Using RR instead of R+hR+h in the formulas for gravitational potential and gravitational acceleration is incorrect because the point is above Earth’s surface. Always use the distance from Earth’s centre, which is R+hR+h.

  • Taking the ratio of the equations incorrectly can lead to algebra errors. Here, dividing ∣V∣|V| by g′g' gives R+hR+h, not hh directly. First find R+hR+h, then subtract RR.

  • Forgetting unit conversion between metres and kilometres gives a wrong final answer. The calculation gives 1.6×106 m1.6 \times 10^6 \, \text{m}, which must be converted to 1600 km1600 \, \text{km}.

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