मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

Is there any meaning of “weight of the earth”?

Advertisements
Advertisements

प्रश्न

Is there any meaning of “weight of the earth”?

दीर्घउत्तर
Advertisements

उत्तर

The weight of a body is always due to its gravitational attraction to earth. As law of gravitational attraction is universal so it applies to any two bodies (earth and sun as well). So we can define the weight of earth with respect to a body of mass comparable to or greater than earth’s, due to the gravitational attraction between earth and that body. But practically nobody on earth has mass comparable to earth, so the weight of earth will be a meaningless concept with respect to the earth frame.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Gravitation - Short Answers [पृष्ठ २२३]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 11 Gravitation
Short Answers | Q 2 | पृष्ठ २२३

संबंधित प्रश्‍न

If heavier bodies are attracted more strongly by the earth, why don't they fall faster than the lighter bodies?


An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration g. Who exerts the force needed to accelerate the earth with this acceleration g?


The acceleration of the moon just before it strikes the earth in the previous question is


Suppose, the acceleration due to gravity at the earth's surface is 10 m s−2 and at the surface of Mars it is 4⋅0 m s−2. A 60 kg passenger goes from the earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of the following figure best represents the weight (net gravitational force) of the passenger as a function of time?


Find the acceleration due to gravity of the moon at a point 1000 km above the moon's surface. The mass of the moon is 7.4 × 1022 kg and its radius is 1740 km.


What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of 8848 m. The value at sea level is 9.80 m s−2.


Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 m s−2. The radius of the earth is 6400 km.


A particle is fired vertically upward from earth's surface and it goes up to a maximum height of 6400 km. Find the initial speed of particle.


A particle is fired vertically upward with a speed of 15 km s−1. With what speed will it move in interstellar space. Assume only earth's gravitational field.


A mass of 6 × 1024 kg (equal to the mass of the earth) is to be compressed in a sphere in such a way that the escape velocity from its surface is 3 × 108 m s−1. What should be the radius of the sphere?


Explain the variation of g with latitude.


Explain the variation of g with altitude.


Calculate the change in g value in your district of Tamil nadu. (Hint: Get the latitude of your district of Tamil nadu from Google). What is the difference in g values at Chennai and Kanyakumari?


The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity ______.


Which of the following options are correct?

  1. Acceleration due to gravity decreases with increasing altitude.
  2. Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).
  3. Acceleration due to gravity increases with increasing latitude.
  4. Acceleration due to gravity is independent of the mass of the earth.

A pebble is thrown vertically upwards from the bridge with an initial velocity of 4.9 m/s. It strikes the water after 2 s. If acceleration due to gravity is 9.8 m/s2. The height of the bridge and velocity with which the pebble strikes the water will respectively be ______.


If the radius of the earth shrinks by 2% while its mass remains the same. The acceleration due to gravity on the earth's surface will approximately ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×