Physics Gravitation

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New answer posted

7 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

(a) Decreases - Acceleration due to gravity at depth h is given by gh = (1 – 2hRe )g, where Re=Radiusoftheearth, g = acceleration due to gravity on the surface of the Earth. From this equation, it is clear that acceleration due to gravity decreases with increase in height

 

(b) Decreases – Acceleration due to gravity at depth d is given by gd = (1- dRe )g. So the acceleration due to gravity decreases with increase in depth.

 

(c) Mass of the body – Acceleration due to gravity of body mass m is given by the relation g = GMR2 , where G = Universal gravitation constant, M = mass of the Earth and R = radius of the Ear

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New answer posted

8 months ago

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Pallavi Pathak

Contributor-Level 10

8.12

Mass of the Sun, Ms = 2*1030 kg, Mass of the Earth, Me = 6*1024 kg

Orbital radius, r = 1.5*1011 m

Let the mass of the rocket be, m

Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.

From Newton's law of gravitation, we can equate gravitational forces acting on the satellite P under the influence of the Sun and the Earth as:

GmMs(r-x)2 = GmMex2 or( r-xx)2 

MsMer-xx)2 = 2*10306*1024 ,

 r-xx = 577.35 , r = 578.35x ,

x = 1.5*1011578.35 = 2.59 *108 m

New answer posted

8 months ago

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P
Pallavi Pathak

Contributor-Level 10

Ans.8.1

(a) No. Unlike electrical forces, gravitational force is independent of the status of the objects.

 

(b) Yes, the size of the space station is large enough and the astronaut will detect the change in Earth's gravity.

 

(c) Tidal effect depends inversely upon the cube of the distance while gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon's pull is greater than the tidal effect of the Sun's pull.

New answer posted

8 months ago

0 Follower 3 Views

P
Pallavi Pathak

Contributor-Level 10

When there is no normal force acting on a body, weightlessness is experienced, and the body free fall under gravity. The concept of gravity is still acting on the body, however, there is no weight sensation of any reaction force from any surface or ground. The weightlessness felt by the astronauts in an orbiting spacecraft is not because gravity is not present but because while they are moving forward, both, they and the spacecraft are in continuous free fall towards the Earth. The role of the normal force in defining weight can be understood from this phenomenon also how it is different from the actual gravitational force.

New answer posted

8 months ago

0 Follower 5 Views

P
Pallavi Pathak

Contributor-Level 10

When there is no normal force acting on a body, weightlessness is experienced, and the body free fall under gravity. The concept of gravity is still acting on the body, however, there is no weight sensation of any reaction force from any surface or ground. The weightlessness felt by the astronauts in an orbiting spacecraft is not because gravity is not present but because while they are moving forward, both, they and the spacecraft are in continuous free fall towards the Earth. The role of the normal force in defining weight can be understood from this phenomenon also how it is different from the actual gravitational force.

New answer posted

9 months ago

0 Follower 6 Views

P
Pallavi Pathak

Contributor-Level 10

It is due to a concept taken from Newton's Law of Gravitation. It says that each mass element of the shell puts an attractive force on a particle inside but due to the fact that the shape of the shell is symmetrical, these forces cancel out in every direction.
To prove it mathematically, one can use Gauss's Law for Gravitation. It states that inside a uniformly distributed spherical shell, the net gravitational field is zero. It differs from electrostatics, which states that a conducting shell blocks or stops the external electric fields. The shielding concept is not part of gravitation where the external bodies exert a gravitational fo

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New answer posted

9 months ago

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Pallavi Pathak

Contributor-Level 10

Gravitational Potential Energy (U) is the energy an object has due to its position in a gravitational field. It is defined as:

Where M represents the mass of the larger body, m is the mass of the smaller body, r is the distance between them, and G stands for the gravitational constant.
G is written with a negative sign to show that the gravitational potential energy is always lower than zero and the values even decrease as the objects move apart from each other. It implies that to separate two masses, work should be done against the gravity. At infinity, U = 0, implying that no external force is required to keep them apart. When two obj

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