Advertisements
Advertisements
Question
Take the effect of bulging of earth and its rotation in account. Consider the following statements :
(A) There are points outside the earth where the value of g is equal to its value at the equator.
(B) There are points outside the earth where the value of g is equal to its value at the poles.
Options
Both A and B are correct.
A is correct but B is wrong.
B is correct but A is wrong.
Both A and B are wrong.
Solution
A is correct but B is wrong.
We know that the value of acceleration due to gravity decreases when we go up from the surface of the Earth. If we take the into account the effect of bulging of the Earth due to its rotation, we can say that acceleration due to gravity is maximum at the poles and minimum at the equator
So, there are points above both the poles where the value of g is equal to its value at the equator.
APPEARS IN
RELATED QUESTIONS
Is there any meaning of "Weight of the earth"?
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 moon with respect to earth is 0⋅0027 m s−2 and the acceleration of an apple falling on earth' surface is about 10 m s−2. Assume that the radius of the moon is one fourth of the earth's radius. If the moon is stopped for an instant and then released, it will fall towards the earth. The initial acceleration of the moon towards the earth will be
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?
If the acceleration due to gravity at the surface of the earth is g, the work done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is
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.
A body is weighed by a spring balance to be 1.000 kg at the North Pole. How much will it weigh at the equator? Account for the earth's rotation only.
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.
Suppose we go 200 km above and below the surface of the Earth, what are the g values at these two points? In which case, is the value of g small?
A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other objects in the sky and take acceleration due to gravity on the surface of the Earth and Mars as 10 m/s2 and 4 m/s2 respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.
If R is the radius of the earth and g is the acceleration due to gravity on the earth's surface, the mean density of the earth is ______.
A ball is immersed in water kept in container and released. At the same time container is accelerated in horizontal direction with acceleration, `sqrt44` m/s2. Acceleration of ball w.r.t. container is ______ m/s2 (specific gravity of ball = 12/17, g = 10 m/s2)
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 ______.
