The correct graph representing the variation of total energy (ET), kinetic energy (EK) and potential energy (Ep) of a satellite with its distance (r) from the centre of earth is

Question

The correct graph representing the variation of total energy (ET), kinetic energy (EK) and potential energy (Ep) of a satellite with its distance (r) from the centre of earth is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ET=-(GM m/2r); Thus ET increases with increase in r Ek=(GM m/2r) ; Thus Ek decreases with increase in r Ep=(-GM m/r) ; Thus Ep increases with increase in r

Q. The correct graph representing the variation of total energy $(E_{T})$ , kinetic energy $(E_{K})$ and potential energy $(E_{p})$ of a satellite with its distance $(r)$ from the centre of earth is

$E_{T} = - \frac{GM m}{2 r};$
Thus $E_{T}$ increases with increase in $r$
$E_{k} = \frac{GM m}{2 r}$ ; Thus $E_{k}$ decreases with increase in $r$
$E_{p} = \frac{- GM m}{r}$ ; Thus $E_{p}$ increases with increase in $r$

Q. The correct graph representing the variation of total energy (ET​), kinetic energy (EK​) and potential energy (Ep​) of a satellite with its distance (r) from the centre of earth is

ET​=−2rGMm​; Thus ET​ increases with increase in r Ek​=2rGMm​ ; Thus Ek​ decreases with increase in r Ep​=r−GMm​ ; Thus Ep​ increases with increase in r

Q. The correct graph representing the variation of total energy $(E_{T})$ , kinetic energy $(E_{K})$ and potential energy $(E_{p})$ of a satellite with its distance $(r)$ from the centre of earth is

$E_{T} = - \frac{GM m}{2 r};$
Thus $E_{T}$ increases with increase in $r$
$E_{k} = \frac{GM m}{2 r}$ ; Thus $E_{k}$ decreases with increase in $r$
$E_{p} = \frac{- GM m}{r}$ ; Thus $E_{p}$ increases with increase in $r$