The intensity of gravitational field at a point situated at a distance of 8000 km from the centre of the earth is 6N / kg . The gravitational potential at that point is – (in Joule / kg)

The correct option is 4.8 x 10⁷.

Given Data:
Gravitational field intensity ($I$) = $6\text{N/kg}$
Distance from the center of the Earth ($r$) = $8000\text{km}=8000\times {10}^3\text{m}=8\times {10}^6\text{m}$

Formulae and Relation:
The intensity of the gravitational field at a distance $r$ from the center of a planet of mass $M$ is expressed as:
$I=\frac{GM}{r^2}$
where $G$ is the universal gravitational constant.

The gravitational potential ($V$) at the same point is given by:
$V=-\frac{GM}{r}$

By relating these two equations, we get:
$V=-I\times r$

Step-by-step Calculation:
Substitute the given values into the relation:
$V=-6\times (8\times {10}^6)$
$V=-48\times {10}^6\text{J/kg}$
$V=-4.8\times {10}^7\text{J/kg}$

Thus, the magnitude of the gravitational potential at the given point is $4.8\times {10}^7\text{J/kg}$.