Does the Acceleration due to gravity increase going to depth of the Earth?

Answer 1

According to the formulae the Acceleration due to gravity increases overall as we go deep into the Earth since the mass above is being pulled in the opposite direction. However it also increases a little due to the presence of denser material in mantle.

Answer 2

I'm sure that you and I are both going to carefully read and thoroughly comprehend

the entire Answer #1 below, no matter how long it takes. I know I can't wait to

get started.

But before we do, I thought I'd just slip in here and answer your question:

Strange as it may seem, as you descend below the Earth's surface, like in a deep

mine or at the bottom of the ocean, the acceleration of gravity becomes less.

OK. Now let's both start reading:

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Acceleration due to Gravity 'g' Bodies allowed to fall freely were found to fall at the same rate irrespective of their masses (air resistance being negligible). The velocity of a freely falling body increased at a steady rate i.e., the body had acceleration. This acceleration is called acceleration due to gravity - 'g'. We know, F = mg --------------- (2) Where F is the force, m is the mass of the body, g is the acceleration due to gravity, M is the mass of the Earth, R is the radius of the Earth and G is the gravitational constant. From equations (2) and (3), 'g' varies with (a) altitude (b) depth (c) latitude Variation of 'g' with altitude Variation of 'g' with altitude Let a body of mass m be placed on the surface of the Earth, whose mass is M and radius is R. From equation (4) Let the body be now placed at a height h above the Earth's surface. Let the acceleration due to gravity at that position be g|. For comparison, the ratio between g| and g is taken By binomial theorem, h is assumed to be very small when compared to radius R of the Earth. Hence, they can be neglected This shows that acceleration due to gravity decreases with increase in altitude. Loss in weight at height h(h< From equation (7) __________(7) Variation of 'g' with depth Variation of 'g' with depthConsider a body of mass m, lying on the surface of the Earth of radius R and mass M. Let g be the acceleration due to gravity at that place. Let the body be taken to a depth d from the surface of the Earth. Then, the force due to gravity acting on this body is only due to the sphere of radius R. (R - d). If g| is the acceleration due to gravity at depth 'd' Let the Earth be of uniform density r and its shape be a perfect sphere. (Where r is the density of the Earth) Comparing g| and g The acceleration due to gravity decreases with increase in depth. If d = R, then g| = 0. Weight of a body at the centre of the Earth is zero. Variation of 'g' with latitude The value of g changes from place to place due to the elliptical shape of the Earth and the rotation of the Earth. Due to the shape of the Earth, From equation (4) Hence, it is inversely proportional to the square of the radius. It is least at the equator and maximum at the poles, since the equatorial radius (6378.2 km) is more than the polar radius (6356.8 km) Due to the rotation of the Earth If 'w' is the angular velocity of the Earth and f is the latitude of the place, Circular motion Suppose a body is undergoing circular motion about a circle of radius r, with an angular velocity w rad/s as shown in the figure. From geometry, we can observe that the corresponding distance travelled in 1 second is equal to rw. Since the distance travelled in 1 second is velocity, v = rw. Every body undergoing circular motion with a constant angular velocity is said to be undergoing uniform circular motion. It experiences acceleration towards the centre of the circle of Since it is directed towards the centre of the circle, it is called centripetal acceleration. Therefore, centripetal acceleration is associated with uniform circular motion and directed towards the centre of the circle Let us now consider earth as a sphere of radius R, undergoing uniform circular motion about its polar axis, connecting the north and south poles. Equator is the horizontal circle passing through the centre of this axis, P1. What is latitude? Every point on the sphere lies on the same latitude, which lie on the base of the cone whose axis coincides with the polar axis and whose generators make an angle f with the horizontal or equatorial plane. The angle f is called the latitude of the place. Latitude of equator = 0o and latitude of north pole = 90o and latitude of south pole = -90o. We can observe that if the Earth is considered to be a sphere of radius R, the radius of the smaller circle in the latitude f=Rcosf. A particle on the latitude f which is undergoing uniform circular motion with angular velocity 'w', experiences centripetal acceleration directed towards the centre of the small circle OI. This acceleration 'a' can be resolved into two components, tangential and vertical. Gravitational acceleration 'g' acts on the body. Since all the forces acting on the body at the latitude f result in uniform circular motion, the net of all forces should be equal to centripetal force. At the equator, At the pole, Hence, the gravitation acceleration is maximum at the poles and minimum at the equator. We should observe that the net of all the forces acting on the body results in uniform circular motion, which means that uniform circular motion is the result of all the forces acting on the body. At the equator f = 0 At the poles, f = 90o, cos f = 0 It is less at the equator and maximum at the poles.

Answer 3

The deeper one is the more weight above them exerting a force on them.