Question

Mass and the radius of the moon are 7.2 *10 power 22 kg and 1.7 *10power 3 km respectivelycalculate the acceleration due to gravity of the moon and the weight of a man of mass of 70 kg on the moon?

Answer 1

General Expression of acceleration on a planet/moon is :

$\rm \: a = \dfrac{Gm_{p} }{ {r}^{2} }$

• m_(p) is mass of planet or moon.

$\rm \implies \: a = \dfrac{(6.6 \times {10}^{ - 11} )\times(7.2 \times {10}^{22}) }{ {(1.7 \times {10}^{3} \times 1000) }^{2} }$

$\rm \implies \: a = \dfrac{(6.6 \times {10}^{ - 11} )\times(7.2 \times {10}^{22}) }{ {(1.7 \times {10}^{6} ) }^{2} }$

$\rm \implies \: a = \dfrac{47.52 \times {10}^{ 11} }{ 2.89 \times {10}^{12} }$

$\boxed{ \rm \implies \: a = 1.64 \: m {s}^{ - 2} }$

Now, weight of 70 kg man:

$\rm W = m \times a$

$\implies \rm W = 70\times 1.64$

$\boxed{\implies \rm W = 114.8 \: N}$

Answer 2

Explanation:

General Expression of acceleration on a planet/moon is :

\rm \: a = \dfrac{Gm_{p} }{ {r}^{2} } a=

r

2

Gm

p

m_(p) is mass of planet or moon.

\rm \implies \: a = \dfrac{(6.6 \times {10}^{ - 11} )\times(7.2 \times {10}^{22}) }{ {(1.7 \times {10}^{3} \times 1000) }^{2} } ⟹a=

(1.7×10

3

×1000)

2

(6.6×10

−11

)×(7.2×10

22

)

\rm \implies \: a = \dfrac{(6.6 \times {10}^{ - 11} )\times(7.2 \times {10}^{22}) }{ {(1.7 \times {10}^{6} ) }^{2} } ⟹a=

(1.7×10

6

)

2

(6.6×10

−11

)×(7.2×10

22

)

\rm \implies \: a = \dfrac{47.52 \times {10}^{ 11} }{ 2.89 \times {10}^{12} } ⟹a=

2.89×10

12

47.52×10

11

\boxed{ \rm \implies \: a = 1.64 \: m {s}^{ - 2} }

⟹a=1.64ms

−2

Now, weight of 70 kg man:

\rm W = m \times aW=m×a

\implies \rm W = 70\times 1.64⟹W=70×1.64

\boxed{\implies \rm W = 114.8 \: N}

⟹W=114.8N