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Every day the earth advances in the orbit by approximately 1^o. Then, it will have to rotate by 361° (which we define as 1 day) to have sun at zenith point again. Since 361° corresponds to 24 hours; extra 1° corresponds to approximately 4 minute [3 min 59 seconds].

The trajectory of a particle under gravitational force of the earth will be a conic section (for motion outside the earth) with the centre of the earth as a focus. Only (c) meets this requirement.

 The gain in its potential energy is - mgR/2.

Correct answer is (B) zero.

No, for stable orbit the plane of the orbit must pass through the centre (equatorial plane) of the planet.

 When a satellite of mass m revolves in a circular orbit of radius r, around the earth of mass m, the gravitational attraction of earth provides the required centripetal force

i.e., \(\frac{mv^2}{r}\) = \(\frac{GMm}{r^2}\)

or mv²r = GMm

or (mvr)v = a constant

or Lv = a constant ..............(i)

where mvr = L, angular momentum of a satellite.

According to a law of conservation of momentum if no external torque acts on the satellite, then its angular momentum L is conserved. Since, the air friction will provide a retarding torque to satellite, therefore, there will be decrease in angular momentum of the satellite in air. As a result of it, the velocity of satellite increase according to relation (i).

The answer is (D) N

Speed of the car (v) = 10 m/s ; Radius of the path (r) = 10 m 

Weight of the car (m) = 1000kg

Centripetal force required (F_c) =  \(\cfrac{mv^2}{r}= \cfrac{1000\,\times\,10\,\times\,10}{10}\) = 10,000 = 10⁴ N

The required centripetal force is provided by the friction between the tyres of the car and the road.

The force of attraction between moon and earth is given by F = \(\cfrac{GMm}{R^2}\)

M = mass of the earth ; 

m = mass of the moon ; 

R = radius of the earth

Here the gravitational interaction between moon and earth disappears. 

∴ G = 0 ⇒ F = 0

• Therefore the moon neither revolves around the earth nor fall into the earth.
• It takes a straight path away from the earth.

Heavy and light objects, when released from the same height, fall towards the Earth at the same speed., i.e., they have the same acceleration.

F = \(G\frac{m_1m_2}{d^2}\)

Whenever objects fall towards the earth under the gravitational force alone, we say that the objects are in free fall.

Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force. Therefore this acceleration is called the acceleration due to the gravitational force of the earth.

Mass of moon is lesser than that of earth. Due to this the moon exerts lesser force of attraction on objects. Hence weight of an object on the moon 1/6th its weight on the earth.

The motion of a body under the influence of gravity alone is called a free fall. 

The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity. 

It is a vector quantity. 

Its SI unit is ms^-2 .

Variation of acceleration due to gravity: 

(i) Effect of altitude: At a height h,: The value of g decreases with the increase in h. 

(ii) Effect of depth: At a depth d, : The value of g decreases with the increase in depth d and becomes zero at the centre of the earth. 

(iii) Effect of non-sphericity of the earth: The earth has an equatorial bulge and is flattened at the poles. So the value of g is minimum at the equator and maximum at the poles.

Correct option is: (A) T² ∝ R^5/2