Question

An earth like planet has a radius equal to double the earth's radius. The acceleration due to gravity on its surface will be
a) g
b) g/2
c) 2g
d) g2

Plz explain​

Answer 1

b) g/2

(or)

C) 2g

explanation

Answer 2

The acceleration due to gravity of a an earth like planet with its radius double to that of earth's is  $\frac{g}{4}$

Explanation:

1. Acceleration due to gravity is the acceleration gained by an object due to gravitational force. Its SI unit is $m/s^2$.
2. It has both magnitude and direction, hence, it’s a vector quantity.
3. Acceleration due to gravity is represented by 'g'.
4. The standard value of 'g' on the surface of the earth at sea level is $9.8\ m/s^2$.
5. Value of 'g' on any planet of mass 'M' and radius 'r' is given by, $g=\frac{GM}{r^2} \ m/s^2$                                                                                          here, G- Universal gravitational constant $(6.67x10^{-11} \ \ Nm^2/kg^2)$  
6. let the acceleration due to gravity on earth be , $g_e=\frac{GM_e}{r_e^2}$  
7. now for earth like planet we have mass of planet equal to that of earth and radius is double to that of earth's, $M_p=M_e, \ \ \ \ r_p=2r_e$  
8. hence the acceleration due to gravity on this planet is ,                                     [tex]g_p=\frac{GM_p}{r_p^2}\\ g_p=\frac{GM_e}{(2r_e)^2}=\frac{GM_e}{4r_e^2}\\ g_p=\frac{g_e}{4} [/tex]    ------>(Answer)