At what height from the surface of earth the gravitation potential and the value of $g$ are $- 5.4 \times 10^{7} J {k g}^{- 2}$ and $6.0 {m s}^{- 2}$ respectively? Take the radius of earth as $6400 k m$ .

Question

At what height from the surface of earth the gravitation potential and the value of

g

are

- 5.4 \times 10^{7} J {k g}^{- 2}

and

6.0 {m s}^{- 2}

respectively? Take the radius of earth as

6400 k m

.

Toppr Admin · Accepted Answer

V-x2212-GMR-h-x2212-5-4-xD7-107 - -1-g-GM-R-h-2-6 - -2-Dividing the two equations- we get-x21D2-5-4-xD7-107-R-h-6-x21D2-R-h-9000-xA0-km-x21D2-h-2600-xA0-km-xA0-

At what height from the surface of earth the gravitation potential and the value of $g$ are $- 5.4 \times 10^{7} J {k g}^{- 2}$ and $6.0 {m s}^{- 2}$ respectively? Take the radius of earth as $6400 k m$ .
$2600 k m$
$1600 k m$
$1400 k m$
$2000 k m$

$V = - \frac{G M}{R + h} = - 5.4 \times 10^{7}$ ....... (1)

$g = \frac{G M}{(R + h)^{2}} = 6$ ...... (2)

Dividing the two equations, we get:

$\Rightarrow \frac{5.4 \times 10^{7}}{(R + h)} = 6$

$\Rightarrow R + h = 9000 k m$

$\Rightarrow h = 2600 k m$

At what height from the surface of earth the gravitation potential and the value of g are −5.4×107 Jkg−2 and 6.0 ms−2 respectively? Take the radius of earth as 6400 km.2600 km1600 km1400 km2000 km

V=−GMR+h=−5.4×107 ....... (1)g=GM(R+h)2=6 ...... (2)Dividing the two equations, we get:⇒5.4×107(R+h)=6⇒R+h=9000 km⇒h=2600 km

At what height from the surface of earth the gravitation potential and the value of $g$ are $- 5.4 \times 10^{7} J {k g}^{- 2}$ and $6.0 {m s}^{- 2}$ respectively? Take the radius of earth as $6400 k m$ .
$2600 k m$
$1600 k m$
$1400 k m$
$2000 k m$

$V = - \frac{G M}{R + h} = - 5.4 \times 10^{7}$ ....... (1)

$g = \frac{G M}{(R + h)^{2}} = 6$ ...... (2)

Dividing the two equations, we get:

$\Rightarrow \frac{5.4 \times 10^{7}}{(R + h)} = 6$

$\Rightarrow R + h = 9000 k m$

$\Rightarrow h = 2600 k m$