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Standard XII

Physics

Satellites

It has been m...

Question

It has been mentioned in the passage that as $r$ decreases, $E$ decreases but $K$ increases. The increase in $K$ is [ $E =$ total mechanical energy, $r =$ orbital radius, $K =$ kinetic energy]

due to increase in gravitational

P E

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due to decrease in gravitational

P E

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due to work done by air friction force

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both (b) and (c)

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Solution

The correct option is B due to decrease in gravitational $P E$
$P . E = - \frac{G M m}{r}$ and $K . E = \frac{1}{2} m v^{2}$ where $v^{2} \propto \frac{1}{r}$
Now as $r$ decreases then its P.E decreases and also K.E increases
Thus decrease in gravitational P.E helps the K.E to increase.

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It has been mentioned in the passage that as r decreases, E decreases but K increases. The increase in K is [E= total mechanical energy, r= orbital radius, K= kinetic energy]

The correct option is B due to decrease in gravitational PEP.E=−GMmr and K.E=12mv2 where v2∝1rNow as r decreases then its P.E decreases and also K.E increasesThus decrease in gravitational P.E helps the K.E to increase.

It has been mentioned in the passage that as $r$ decreases, $E$ decreases but $K$ increases. The increase in $K$ is [ $E =$ total mechanical energy, $r =$ orbital radius, $K =$ kinetic energy]

The correct option is B due to decrease in gravitational $P E$
$P . E = - \frac{G M m}{r}$ and $K . E = \frac{1}{2} m v^{2}$ where $v^{2} \propto \frac{1}{r}$
Now as $r$ decreases then its P.E decreases and also K.E increases
Thus decrease in gravitational P.E helps the K.E to increase.