Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
The acceleration due to gravity at a point at height h from the surface of the moon is given by \[g = \frac{GM}{r^2}\] ,
where M is the mass of the moon; r is the distance of point from the centre of the moon and G is universal gravitational constant.
\[\therefore g = \frac{GM}{\left( R + h \right)^2}\]
\[ \Rightarrow g = \frac{6 . 67 \times {10}^{- 11} \times 7 . 4 \times {10}^{22}}{\left( 1740 + 1000 \right)^2 \times {10}^6}\]
\[ \Rightarrow g = \frac{6 . 67 \times 7 . 4 \times {10}^{11}}{\left( 1740 + {1000}^2\times {10}^6 \right)}\]
\[ \Rightarrow g = \frac{6 . 67 \times 7 . 4 \times {10}^{11}}{2740 \times 2740 \times {10}^6}\]
\[ \Rightarrow g = 0 . 65 m/ s^2\]
APPEARS IN
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
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
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.
Find the height over the Earth's surface at which the weight of a body becomes half of its value at the surface.
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 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 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.
If the acceleration due to gravity becomes 4 times its original value, then escape speed ____________.
Explain the variation of g with latitude.
Explain the variation of g with altitude.
Explain the variation of g with depth from the Earth’s surface.
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?
If both the mass and the radius of the earth decrease by 1%, then the value of acceleration due to gravity will
Which of the following options are correct?
- Acceleration due to gravity decreases with increasing altitude.
- Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).
- Acceleration due to gravity increases with increasing latitude.
- Acceleration due to gravity is independent of the mass of the earth.
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 ______.
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 ______.
