Question

calculate the value of acceleration due to gravity at point:
(a)5.0 km above the earth`s surface and
(b)5.0 km below the earth`s surface.Radius of earth=6400km
and the value of g at the surface of the earth is 9.80 m/s^2

Answer 1

Answer on Question 74125, Physics, Astronomy, Astrophysics

Question:

Calculate the value of acceleration due to gravity at point:

a) $5.0 \, \text{km}$ above the Earth's surface and

b) $5.0 \, \text{km}$ below the Earth's surface.

Radius of Earth is $6400 \, \text{km}$ and the value of $g$ at the surface of the Earth is $9.8 \, \text{m/s}^2$ .

Solution:

a) As we know, the acceleration due to gravity on the surface of the Earth is given by the formula:

$$g = \frac {G M}{R ^ {2}}, (1)$$

here, $G$ is the universal gravitational constant, $M$ is the mass of the Earth, $R$ is the radius of the Earth.

At the height $h$ above the Earth, the acceleration due to gravity is given by:

$$g _ {h} = \frac {G M}{(R + h) ^ {2}}. (2)$$

Let's divide equation (2) by equation (1):

$$\frac {g _ {h}}{g} = \frac {G M}{(R + h) ^ {2}} \cdot \frac {R ^ {2}}{G M} = \frac {R ^ {2}}{(R + h) ^ {2}} = \frac {1}{\left(1 + \frac {h}{R}\right) ^ {2}} = \left(1 + \frac {h}{R}\right) ^ {- 2} = \left(1 - \frac {2 h}{R}\right).$$

Finally, we get:

$$g _ {h} = g \left(1 - \frac {2 h}{R}\right) = 9. 8 \frac {m}{s ^ {2}} \cdot \left(1 - \frac {2 \cdot 5 \cdot 1 0 ^ {3} m}{6 . 4 \cdot 1 0 ^ {6} m}\right) = 9. 7 8 \frac {m}{s ^ {2}}.$$

b) Let's denote the density of the Earth as $\rho$ . Then, the mass of the Earth can be written as follows:

$$M = \rho V = \frac {4}{3} \pi R ^ {3} \rho ,$$

here, $V = \frac{4}{3}\pi R^3$ is the volume of the Earth.

Then, we can write the acceleration due to gravity at the surface of the Earth:

$$g = \frac {G M}{R ^ {2}} = \frac {G}{R ^ {2}} \cdot \frac {4}{3} \pi R ^ {3} \rho = \frac {4}{3} \pi G R \rho \quad (3)$$

At the depth $d$ below the Earth's surface the acceleration due to gravity is given by:

$$g _ {d} = \frac {4}{3} \pi G (R - d) \rho \quad (4)$$

Let's divide equation (4) by equation (3):

$$\frac {g _ {d}}{g} = \frac {\frac {4}{3} \pi G (R - d) \rho}{\frac {4}{3} \pi G R \rho} = \frac {R - d}{R} = \left(1 - \frac {d}{R}\right).$$

Finally, we get:

$$g _ {d} = g \left(1 - \frac {d}{R}\right) = 9.8 \frac {m}{s ^ {2}} \cdot \left(1 - \frac {5 \cdot 1 0 ^ {3} m}{6 . 4 \cdot 1 0 ^ {6} m}\right) = 9.79 \frac {m}{s ^ {2}}.$$

Answer:

a) $g_{h} = 9.78\frac{m}{s^{2}}.$

b) $g_{d} = 9.79\frac{m}{s^{2}}.$

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