Gravity in equator versus pole of Jupiter

GRAVITY IN EQUATOR V/S POLE OF JUPITER

Jupiter is the fifth planet in our Solar System. Gravity on the surface of Jupiter varies due to its rotation. Since Jupiter rotates, all points on its surface perform circular motion except the North and South Pole. The points at equator perform circular motion of maximum radii as compared to other points. The polar points do not perform circular motion but rotate about its own axis. 

Consider two points, equator and North Pole on Jupiter’s surface. We’ll compare acceleration due to gravity (g) between these two points.

ASSUMPTIONS

1. The equator and pole locations are not terrain but plains

2. Density of Jupiter is constant throughout

3. Jupiter is a perfect homogeneous sphere

4. Angular velocity is constant throughout

5. Effect of gravity of satellites is negligible

PHYSICAL CHARACTERISTICS

Mean Radius R= 69,911 km

Mass M = 1.8986*10²⁷ kg

Average Rotational Time period T = 9 hours 55 minutes 30 second

CALCULATION

Acceleration due to gravity on pole:

g = GM/R²

G – Universal Gravitation constant = 6.67*10^-11 Nm²/kg²

g = [6.67*10^-11 * 1.8986*10²⁷]/ [69911*1000]²

g = 25.9 m/s²

The above formula does not account for rotation of the planet. Hence the answer is gravity at the pole as pole points do not rotate.

g (φ) = 25.9 m/s²

φ – Latitude = 90⁰ for poles

Average Angular velocity:

ω = 2π / T (rad/s)

ω = 2π / [9(60*60) + 55*60 + 30] (rad/s) 

ω = 2π/35730

ω = 1.758518*10^-4 rad/s

Acceleration due to gravity on equator:

We can find acceleration due to gravity on equator by using the formula,

g’(φ) = g(φ) – Rω²cos²φ

φ - Latitude = 0⁰ for equator

g’(φ) = 25.9 – 69911000*(1.758518*10^-4)²

g’(φ) = 25.9 – 2.161917

g’(φ) = 23.7381 m/s²

CONCLUSION

On comparing g (φ) and g’ (φ) we observe that gravity at pole is greater than that at the equator. The difference is g (φ) - g’ (φ) = 25.9 – 23.7381 = 2.1619 m/s²

For a body of mass 50Kg,

Weight at pole = 50* g (φ) = 50*25.9 = 1295 N

Weight at equator = 50* g’ (φ) = 50*23.7381 = 1186.905 N

Thus a body will weigh approximately 108 N more in pole than equator.

Gravity in Chicago versus Quito

GRAVITY IN CHICAGO V/S QUITO

Gravity on the surface of Earth varies due to its rotation. Earth rotates at a speed of 1040 mph or at an angular velocity of 7.27*10^-5 rad/s. Since Earth rotates, all points on its surface perform circular motion except the North and South Pole. The points at equator perform circular motion of maximum radii as compared to other points. The polar points do not perform circular motion but rotate about its own axis.

Consider two cities, Chicago, USA and Quito, Ecuador on Earth’s surface. We’ll compare acceleration due to gravity (g) between these two cities.

ASSUMPTION:

Density of earth is constant throughout

CALCULATION:

The International Gravity Formula [IGF] for Earth is,

g (φ) = 9.780327[1+0.0053024sin²φ-0.0000058sin²2φ]

Chicago: latitude [φ] = 41.8781⁰ N, height above sea level h= 181m

g (φ) = 9.780327[1+0.0053024sin²(41.8781)-0.0000058sin²(83.7562)]

g (φ) = 9.780327[1+0.0053024(0.6675)²-0.0000058(0.9940)²]

g (φ) = 9.780327[1+2.3625*10^-3-5.7306*10^-6]

g (φ) = 9.780327[1.0023682]

g (φ) = 9.80348 m/s²

Since Chicago is not exactly at sea level, we need to use the Free Air Correction [FAC] so as to find acceleration due to gravity at that height and then add it to our original answer

FAC is given by, ∆g_h = -3.086*10^-6*h

∆g_h = -3.086*10^-6*181

∆g_h = -5.58566*10^-4 m/s²

Now g (φ, h) = g (φ) + ∆g_h

g (φ, h) = 9.80348-5.58566*10^-4

g (φ, h) = 9.802921 m/s²

Quito: latitude [φ] = 0.1807⁰ S, height above sea level h = 2850m

g’ (φ) = 9.780327[1+0.0053024sin²φ-0.0000058sin²2φ]

g’ (φ) = 9.780327[1+0.0053024sin²(0.1807)-0.0000058sin²(0.3614)]

g’ (φ) = 9.780327[1+5.274*10^-8-2.3075*10^-10]

g’ (φ) = 9.780327[1.000000053]

g’ (φ) = 9.7803275 m/s²

Since Quito is way above sea level, we need to use the Free Air Correction [FAC] so as to find acceleration due to gravity at that height and then add it to our original answer

FAC is given by, ∆g_h = -3.086*10^-6*h

∆g_h = -3.086*10^-6*2850

∆g_h = -8.7951*10^-3 m/s²

Now g’ (φ, h) = g’ (φ) + ∆g_h

g’ (φ, h) = 9.7803275-8.7951*10^-3

g’ (φ, h) = 9.771532 m/s²

CONCLUSION:

On comparing g (φ, h) and g’ (φ, h) we observe that gravity in Chicago is greater than that in Quito. The difference is g (φ, h) - g’ (φ, h) = 9.802921-9.771532 = 0.031389 m/s²

For a body of mass 50Kg,

Weight in Chicago = 50* g (φ, h) = 50*9.802921 = 490.146 N

Weight in Quito = 50* g’ (φ, h) = 50*9.771532 = 488.576 N 

Thus a person in Chicago would weigh 1.57 N less in Quito.

Gravity in Equator versus Pole of Earth

GRAVITY IN EQUATOR V/S POLE OF EARTH

Gravity on the surface of Earth varies due to its rotation. Earth rotates at a speed of 1040 mph or at an angular velocity of 7.27*10^-5 rad/s. Since Earth rotates, all points on its surface perform circular motion except the North and South Pole. The points at equator perform circular motion of maximum radii as compared to other points. Hence the angular velocity of Earth is max at its equator, decreases as we move away from equator and eventually zero at the poles.

  

Consider two points, equator and North Pole on Earth’s surface. We’ll compare acceleration due to gravity (g) between these two points.

ASSUMPTIONS:

1. The equator and pole locations are at sea level

2. Density of earth is constant throughout

CALCULATION:

 

The International Gravity Formula [IGF] for Earth is,

g (φ) = 9.780327[1+0.0053024sin²φ-0.0000058sin²2φ]

 North Pole: latitude [φ] = 90⁰

g (φ) = 9.780327[1+0.0053024sin²90-0.0000058sin²180]

g (φ) = 9.780327[1+0.0053024-0]

g (φ) = 9.780327[1.0053024]

g (φ) = 9.83218 m/s²

Equator: latitude [φ] = 0⁰

g’ (φ) = 9.780327[1+0.0053024sin²φ-0.0000058sin²2φ]

g’ (φ) = 9.780327[1+0.0053024sin²0-0.0000058sin²0]

g’ (φ) = 9.780327[1+0-0]

g’ (φ) = 9.780327 m/s²

  

CONCLUSION:

On comparing g (φ) and g’ (φ) we observe that gravity at pole is greater than that at the equator. The difference is g (φ) - g’ (φ) = 9.83218-9.789327 = 0.051853 m/s²

For a body of mass 50Kg,

Weight at pole = 50* g (φ) = 50*9.83218 = 491.609N

Weight at equator = 50* g’ (φ) = 50*9.789327 = 489.466N