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Gravitation - Acceleration due to Gravity (above and below Earth's surface)

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Acceleration due to gravity (gg) is the acceleration gained by an object due to the gravitational force of the Earth. At the surface, it is given by g=GMR2g = \frac{GM}{R^2}, where GG is the gravitational constant, MM is the mass of Earth, and RR is the radius of Earth.

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Variation with Altitude (Height): As we move above the Earth's surface, the distance from the center of the Earth increases, which causes the value of gg to decrease. For small heights (h≪Rh \ll R), the decrease is linear.

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Variation with Depth: As we go below the Earth's surface, the effective mass of the Earth attracting the object decreases. Consequently, the value of gg decreases linearly with depth dd.

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At the center of the Earth (d=Rd = R), the acceleration due to gravity becomes zero (g=0g = 0).

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The value of gg is maximum at the Earth's surface and decreases whether we go upwards or downwards.

📐Formulae

g=GMR2g = \frac{GM}{R^2}

gh=g(RR+h)2g_h = g \left( \frac{R}{R+h} \right)^2

gh≈g(1−2hR) (valid only if h≪R)g_h \approx g \left( 1 - \frac{2h}{R} \right) \text{ (valid only if } h \ll R)

gd=g(1−dR)g_d = g \left( 1 - \frac{d}{R} \right)

💡Examples

Problem 1:

At what height hh above the Earth's surface will the value of gg be half of its value on the surface? (Take RR as the radius of Earth)

Solution:

Given gh=g2g_h = \frac{g}{2}. Using the exact formula: g2=g(RR+h)2\frac{g}{2} = g \left( \frac{R}{R+h} \right)^2 12=(RR+h)2\frac{1}{2} = \left( \frac{R}{R+h} \right)^2 Taking the square root on both sides: 12=RR+h\frac{1}{\sqrt{2}} = \frac{R}{R+h} R+h=2RR + h = \sqrt{2}R h=(2−1)Rh = (\sqrt{2} - 1)R Substituting 2≈1.414\sqrt{2} \approx 1.414: h≈0.414Rh \approx 0.414R

Explanation:

Since the value of gg reduces significantly (to 50%50\%), we cannot use the approximation formula gh=g(1−2h/R)g_h = g(1 - 2h/R). We must use the general formula to find the height in terms of the Earth's radius.

Problem 2:

Find the depth dd below the Earth's surface where the acceleration due to gravity is 25%25\% of its value on the surface.

Solution:

Given gd=25% of g=g4g_d = 25\% \text{ of } g = \frac{g}{4}. Using the formula for depth: gd=g(1−dR)g_d = g \left( 1 - \frac{d}{R} \right) g4=g(1−dR)\frac{g}{4} = g \left( 1 - \frac{d}{R} \right) 14=1−dR\frac{1}{4} = 1 - \frac{d}{R} dR=1−14\frac{d}{R} = 1 - \frac{1}{4} dR=34\frac{d}{R} = \frac{3}{4} d=34Rd = \frac{3}{4}R

Explanation:

At a depth equal to three-fourths of the Earth's radius, the gravity reduces to one-fourth of its surface value because only the inner core of radius R/4R/4 contributes to the gravitational pull.

Acceleration due to Gravity (above and below Earth's surface) Revision - Class 11 Physics CBSE