A body weighs $9 k g$ on the surface of earth. How much will it weigh on the surface of moon where acceleration due to gravity is one-sixth of its value on earth? What is the mass of body on earth and moon?

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Solution

Step 1: Given
$L e t u s a s s u m e g_{e} \to a c c e l e r a t i o n d u e t o g r a v i t y o n e a r t h A c c e l e r a t i o n d u e t o g r a v i t y o n m o o n, g_{m} = \frac{g_{e}}{6} W e i g h t o f t h e b o d y o n e a r t h, W_{e} = 9 k g f W e i g h t o f t h e b o d y o n m o o n, W_{m} = ? A l s o, M a s s o f b o d y o n e a r t h, M_{e} = ? M a s s o f b o d y o n m o o n, M_{m} = ?$
Step 2: Writing the expressions for weight on earth and moon in terms of mass and acceleration due to gravity
$M a s s o f b o d y i s t h e s a m e e v e r y w h e r e . S o, M_{e} = M_{g} = 9 k g (s a y M) N o w, W_{e} = M \times g_{e} W_{m} = M \times g_{m} = M \times \frac{g_{e}}{6}$
Step 3: Comparing the weights to find out the acceleration due to gravity on moon
$\frac{W_{e}}{W_{m}} = \frac{M \times g_{e}}{M \times g_{m}} = \frac{g_{e}}{g_{m}} = \frac{g_{e}}{\frac{g_{e}}{6}} = 6$
Step 4: Substituting the given values to find the value of $W_{m}$ .
$\frac{W_{e}}{W_{m}} = \frac{g_{e}}{g_{m}} \frac{9}{W_{m}} = 6 W_{m} = \frac{9}{6} = 1.5 k g f$
Hence, the weight of the body comes out to be $1.5 k g f$ .

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A body weighs 9kg on the surface of earth. How much will it weigh on the surface of moon where acceleration due to gravity is one-sixth of its value on earth? What is the mass of body on earth and moon?

A body weighs $9 k g$ on the surface of earth. How much will it weigh on the surface of moon where acceleration due to gravity is one-sixth of its value on earth? What is the mass of body on earth and moon?