Gravitation: Important Questions

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Very Short Answer Questions [1 Marks Question]

Ques. How does the Earth keep the majority of its atmosphere?

Ans. The gravitational pull of the Earth keeps it in place. Earth's gravity would be too weak to support a big atmosphere if it were a much smaller planet like Mercury or Pluto.

Ques. What Causes Negative Gravitational Potential Energy?

Ans. This is because, ‘r’ is modest and ‘U’ takes on a big negative value when we are close to the gravitational object M. As the object is moved further away from M, this value decreases from a high negative value to a small negative value until it ultimately hits zero at an infinite distance. As a result, gravity's potential energy is always negative.

Ques. Define Mean solar day.

Ans. The mean solar day is the average of all solar days in which the earth completes one revolution around the sun. The average solar day is divided into twenty-four equal parts, each of which is referred to as an hour.

Ques. What is an artificial satellite's orbital period of rotation in a geostationary orbit?

Ans. To be geostationary on the equatorial plane, the satellite's time period must be equal to the Earth's rotation time period, which is 24 hours.

Ques. Why are space rockets being launched eastward?

Ans. Due to the spinning speed of the Earth, launching a rocket from the east coast provides an additional boost to the rocket. Furthermore, because these rockets head eastward, if something goes wrong during their ascent, the debris will fall into ocean waters, far from densely inhabited areas.

Ques. What are the two variables that determine whether or not a planet has an atmosphere?

Ans. The two variables are the planet's acceleration owing to gravity and the temperature of its surface. Because the temperature affects the speed of gas molecules.

Ques. Why is a body in the centre of the Earth weightless?

Ans. The acceleration due to gravity decreases as the distance from the earth's centre decreases until it reaches zero at the earth's centre.

Ques. What is the difference between a heavier and a lighter object's escape velocity?

Ans. The mass of the body, as well as the direction of projection of the body, have no bearing on escape velocity. The mass and radius of the planet or Earth from which the body is to be projected are the sole factors.

Ques. Is it possible to develop inertia without gaining weight?

Ans. Yes, Inertia, or we can say mass can exist in a body but not weight. Although a body has mass, the gravitational force acting on it can be zero.

Read More: Difference Between Gravitation and Gravity

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Short Answer Questions [2 Marks Question]

Ques. What Is Kepler's Periodic Law?

Ans. The square of a planet's time period of revolution around the sun in an elliptical orbit is precisely proportional to the cube of its semi-major axis, according to Kepler's law of periods.

T² ∝ a³

The shorter the planet's orbit around the sun, the faster it takes to complete one revolution. Kepler's third law is more generalised using the equations of Newton's law of gravitation and laws of motion.

P² = 4π² /[G(M1+ M2)] × a³

Ques. Define sidereal day

Ans. The sidereal day is the interval between two transits of the First Point of Aries in a row. It indicates the time it takes the earth to rotate on its axis in relation to the stars, and because of the earth's orbital velocity, it is nearly four minutes shorter than the solar day.

Ques. What are Gravitational Force's Characteristics?

Ans. Characteristics of Gravitational force are given below.

• The inverse square law governs it.
• In nature, it is always attractive.
• It is a force with a long-range. It stretches beyond infinity.
• The gravitational field particle is the gravitational force's field particle.
• It is the weakest force in the universe.
• It is a conservative force because it is a central force.

Ques. Find an expression for Earth's Density in terms of g and G, assuming Earth is a uniform sphere.

Ans. If the Earth is a uniform sphere with a density ρ, the gravitational acceleration will be.

g = GM/R²

As a result, the above equation gives the density of the earth in terms of g and G.

g = G{(4/3)ðR³ρ}/R²

g = G{(4/3)ðRρ

ρ = 3g/4ðGR

Therefore, the density of the earth in terms of g and G is given by the above equation.

Ques. List the two most important requirements for a geostationary satellite.

Ans. The following are the two requirements for a geostationary satellite.

• A satellite's revolution around the planet should have the same period as the earth's rotation around its axis.
• The rotation of satellites should be in the same direction as the rotation of the earth around its axis, i.e. from west to east in an anti-clockwise orientation.

Ques. What is the angle between a polar satellite's equatorial plane and its orbital plane?

Ans. A polar satellite's equatorial plane and orbital plane are 90 degrees apart, while a geostationary satellite's equatorial and orbital planes are 0 degrees apart.

Ques.Why don't all the molecules fall into the ground like an apple does when it falls from a tree?

Ans. Like an apple falling vertically from a branch, air molecules in the atmosphere experience a vertically downward force due to gravity. The air molecules move at random due to temperature. The velocity of air molecules is not exactly in the vertical downward direction as a result of this. The downward gravitational attraction on ait molecules causes the density of air in the atmosphere near the ground to rise. Because of this, the density of air decreases as we ascend.

Read More: Relation Between G and g

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Long Answer Questions [3 Marks Question]

Ques. Define the escape velocity in terms of the earth. In the mathematical formula for escape velocity, what do R and g stand for?

Ans. The escape velocity is the minimum velocity required for an object to escape the gravitational field of the earth.

V_e = 2gR

R is the radius of the earth, and g denotes the gravitational constant, or acceleration owing to gravity, in the mathematical formula for escape velocity.

Ques. Assuming the globe is a spherical with constant mass density, how much would a body weighing 250N on the surface weigh half way down to the planet's centre?

Ans. At the Earth's surface, the weight of a body with mass m is given by W = mg

Let the depth where the body is located be ‘d’ then d = 1/2 x R where R is the radius of the earth.

Acceleration due to gravity at the depth d will be g_d = (1 - d/R) x g

g_d  = (1 -R/2R) x g = 1/2 x g

Then the weight of the body at depth d will be Wd = mg

= m x 1/2 x g = ½ x mg = 1/2 W

Then W_d = ½ x 250 = 125 N

The weight of a body half-way down to the earth's core is 125N.

Ques. An item of mass m that has been elevated from the earth's surface to a height equal to the radius of the earth. It was taken from a distance of R to 2R from the earth's centre. What will be the increase in the potential energy of the object?

Ans. When an object with mass m is lifted from the earth's surface to a height equal to the radius of the earth, i.e. from R to 2R,

P.E. of the body on the surface of the earth is given by E = -GmM/R

P.E. of the object at an equal to the radius of the earth will be â_< E’ = -GmM/2R

The gain in Potential Energy = E’ - E

= -GmM/2R - (-GmM/R)

= GmM(1/2-1)/R

= GmM/2R Since GM = gR²

= gR²m/2R

= ½(mgR)

Ques. On the surface of the earth, a body weighs 63N. At a height equal to half the earth's radius, what is the gravitational force exerted on it by the earth?

Ans. Let the Weight of the body is W and radius of earth is R.

Then W = 63N

Acceleration due to gravity at height ‘h’ from the surface of earth is gh = g/(1 + h/R)²

For h=R/2 then gh = g/(1+1/2)²

gh  = 4g/9

Weight of body of mass m will be W’ = mgh

= m x 4g/9

W’ = W x 4/9

W’ = 63 x 4/9

W’ = 28N

Thus, at a height equal to half the earth's radius, the gravitational force on the body owing to the earth is 28N.

Read More: Gravitational potential energy

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Very Long Answer Questions [5 Marks Question]

Ques. Give a definition for Gravitational Potential Energy. As a result, derive an expression for the Gravitational Potential Energy of a Body Near the Earth's Surface.

Ans. Gravitational potential energy is the effort done in getting a body from infinity to a point in a gravitational field.

When an object is at a distance from the earth's centre, there is a force of attraction between the earth and the object.

F = GmM/x²

Consider the dW, which is the minimal amount of work required to transport the body through the very short distance dx without accelerating.

dW=Fdx

And the total work done to get the body from infinity to point p, which is r miles from the earth's core.

W(r) = rGmM/x² = -GMm/r 

Hence, the gravitational potential energy is -GMm/r

Ques. What is the weight of a quintal body brought to a height of 1600km above sea level if the radius of the earth is 6400km?

Ans. Let Radius of earth be denoted by R

Then R = 6400km = 6400 x 10³m

Weight of body w=mg

W = mg 

and 1 quintal = 100 kg

Then weight will be mg = 100 x 9.8 N

Weight at height h will be w = mg’

Where g’= g [ R/(R+h) ]²

Now Weight w = mg’

W = mxg [ R/(R+h) ]²

=100 x 9.8 [6400/(6400+1600)]²

W = 64 x 9.8N

W = 627.9 N

Thus the weight of a quintal body is 627.9 N at the height of 1600 km from the surface of the earth.

Ques. On a horizontal table, two heavy spheres, each with a mass of 100kg and a radius of 0.10m, are set 1.0m apart. What is the gravitational force and potential at the intersection of the two spheres' centres? Is it possible for an object to be in equilibrium at that location? Is the equilibrium stable or unstable, if so?

Ans. Given that the mass of each sphere is 100 kg and separation between them is 1 m.

X is the location where the two spheres meet in the middle. At point X, the gravitational force will be zero. This is by the fact that each sphere's gravitational force will act in different directions.

At point X, the gravitational potential exists equal to -GM/(r/2) - GM/(r/2)

= -4GM/r

P.E. = -4GM/r

P.E. = (-4 x 6.67 x 10^-11 x 100)/r

P.E. = -2.67 x 10^-8 J/Kg

Any object placed at any point X will be in a state of balance, but the equilibrium will be unstable. This is due to the fact that any change in the object's position will change the effective force in that direction.

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