Answer
64.8k+ views
Hint: Think about Newton's law of gravitation and try to use the relation between acceleration due to gravity and the universal gravitational constant. The acceleration due to gravity depends on the radius of the planet or satellite and so that has different values for moon and earth as their radii vary.
Complete step by step solution:
The law of universal gravity of Newton is normally that any particle attracts every other particle in the universe with a force strictly proportional and inversely proportional to the distance between its cores. Or, the rule says that every point mass is attracted by force along the line between the two lines, by every other point mass. The power is proportional to the production of the two masses and inversely proportional to their distance.
Now, according to gravity equation, we know that
\[g \propto \dfrac{{GM}}{{{R^2}}}\]
$g$ is acceleration due to gravity,
$G$ is the gravitational constant,
$M$ is the mass of the body and,
$R$ is the radius of the body.
Now, the acceleration due to gravity ${g_p} $ on different planet can be written as
${g_p} = (\dfrac{{{M_p}}}{{{M_e}}} \times {(\frac{{{R_e}}}{{{R_p}}})^2})g$
${M_p} $ is the mass of the object on planet
${M_e} $ is the mass of the object on earth.
${R_p} $ is the radius of the object on the planet.
${R_e} $ is the radius of the object on earth.
Putting the values given in the question,
$ \Rightarrow {g_p} = 2 \times {(\dfrac{1}{2})^2}g$
${g_p} = \dfrac{g}{2}$
${g_p} = 4.9m{s^ {- 2}} $
The acceleration due to gravity ${g_p} $ on different planets is $4.9m{s^ {- 2}} $.
Hence the correct option is A.
Note: Since the force of gravity is directly commensurate with the mass of all interacting objects, more large objects are drawn with a greater force. If the mass of all objects increases, the gravitational force also increases between them. If the mass of one object is doubled, the gravitational power is doubled between them. If the mass of one object is multiplied, the force of gravity is tripled between the objects. The force of gravity on these two objects is quadrupled by the density of all objects, etc.
As the gravitational power is inversely equal to the dividing distance square between the two interacting objects, further dividing distance produces weaker gravitational force. The force of gravitational interaction between them also reduces, when two objects are isolated from each other.
Complete step by step solution:
The law of universal gravity of Newton is normally that any particle attracts every other particle in the universe with a force strictly proportional and inversely proportional to the distance between its cores. Or, the rule says that every point mass is attracted by force along the line between the two lines, by every other point mass. The power is proportional to the production of the two masses and inversely proportional to their distance.
Now, according to gravity equation, we know that
\[g \propto \dfrac{{GM}}{{{R^2}}}\]
$g$ is acceleration due to gravity,
$G$ is the gravitational constant,
$M$ is the mass of the body and,
$R$ is the radius of the body.
Now, the acceleration due to gravity ${g_p} $ on different planet can be written as
${g_p} = (\dfrac{{{M_p}}}{{{M_e}}} \times {(\frac{{{R_e}}}{{{R_p}}})^2})g$
${M_p} $ is the mass of the object on planet
${M_e} $ is the mass of the object on earth.
${R_p} $ is the radius of the object on the planet.
${R_e} $ is the radius of the object on earth.
Putting the values given in the question,
$ \Rightarrow {g_p} = 2 \times {(\dfrac{1}{2})^2}g$
${g_p} = \dfrac{g}{2}$
${g_p} = 4.9m{s^ {- 2}} $
The acceleration due to gravity ${g_p} $ on different planets is $4.9m{s^ {- 2}} $.
Hence the correct option is A.
Note: Since the force of gravity is directly commensurate with the mass of all interacting objects, more large objects are drawn with a greater force. If the mass of all objects increases, the gravitational force also increases between them. If the mass of one object is doubled, the gravitational power is doubled between them. If the mass of one object is multiplied, the force of gravity is tripled between the objects. The force of gravity on these two objects is quadrupled by the density of all objects, etc.
As the gravitational power is inversely equal to the dividing distance square between the two interacting objects, further dividing distance produces weaker gravitational force. The force of gravitational interaction between them also reduces, when two objects are isolated from each other.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)