Question

calculate the amount of work done to move 1 kg mass from the surface of the earth to a point 10⁵ km from the centre of the earth.​

Answer 1

Answer:

The amount of work done will be equal to 5.85×10⁷Joules to move 1Kg mass.

Explanation:

We know that potential energy $P.E.=-\frac{GMm}{R}$

Where G is gravitational constant, $G=6.67*10^{-11}Nm^{2}Kg^{-2}$

Mass of the earth, $M=6*10^{24}Kg$

The radius of the earth, $R=6400Km$

Mass of object, $m=1kg$

Now the mass 1kg has to move from the surface of earth.

The distance between the center of earth O and point B, $r=10^{5}Km$

Consider that $P.E_{A}$ is potential energy of mass at point A on the surface of earth and $P.E._{B}$ is potential energy at point B.

From the work energy theorem:

Work done = increase in potential energy

Work done $= P.E_{B}-P.E_{A}$

$=-\frac{GMm}{r} -(-\frac{GMm}{R} )$

$=GMm(\frac{1}{R} -\frac{1}{r} )$      

$=6.67*10^{-11}Nm^{2}Kg^{-2}*6*10^{24}kg*1kg(\frac{1}{6400km}-\frac{1}{10^{5}km} )$

$=40.02*10^{13}(15.625*10^{-5}-10^{-5})$

$=(585.2*10^{13}Nm^{2})*10^{-5}Km^{-1}$

$=(585.2*10^{13}Nm^{2})*10^{-5}*10^{-3}m^{-1}\\=(585.2*10^{13}Nm^{2})*10^{-8}m^{-1}\\=585.2*10^{5}Nm\\=5.85*10^{7}J$

Therefore, the work done is equal to 5.85×10⁷J.  

   

Answer 2

Answer:

The amount of work done will be equal to 5.85×10⁷Joules to move 1Kg mass.

Explanation:

Given:

Mass of object = 1kg

Distance = 10⁵ km

               =  (10⁵ × 10³) m

              = 10⁸ m

We know that,

Potential energy (P.E) = $-\frac{GMm}{R}$

Where G ⇒ Gravitational constant

M ⇒ Mass of Earth = 6 × 10²⁴

m ⇒ Mass of object = 1 kg

R ⇒ Radius of Earth = 6.4 × 10⁶ m

Now, the distance between point B and the earth's centre O, r = 10⁸ m

Let the potential energy of mass at point A on the surface of earth to be $P.E_{A}$ = $\frac{-GMm}{R}$ and the potential energy at point B $P.E_{B}$ = $- \frac{GMm}{r}$

Therefore, from the work energy theorem,

Work done = increase in potential energy

$W.D = P.E_{B}-P.E_{A}$

$W.D =- \frac{GMm}{r}-(\frac{-GMm}{R})$

$W.D = \frac{GMm}{R}-\frac{GMm}{r}$

$W.D = GMm (\frac{1}{R}-\frac{1}{r})$

$W.D = 6.67$ × $10^{-11}$ × $6$ × $10^{24}$ × $1$ × $(\frac{1}{6.4* 10^{6}} - \frac{1}{10^{8} })$

$W.D=5.85$ × $10^{7} J$

∴ The amount of work done will be equal to 5.85×10⁷Joules to move 1Kg mass.

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