Question

the value of g on the surface of an imaginary planet of mass 1multiply 10 ki power 24kg and radius 2multiply 10 ki power 6​

Answer 1

16.65 m/s² is the required answer.

GIVEN:

• Mass of the planet, M = 1 × 10²⁴ kg

• Radius of the planet , R = 2 × 10⁶ m

TO FIND:

Acceleration due to gravity, g

FORMULA:

g = GM/R²

Where G is the universal gravitational constant.

G = 6.67 × 10^-11 Nm²/kg²

SOLUTION:

g = GM/R²

$g = \dfrac{6.67 \times {10}^{ - 11} \times 1 \times {10}^{24} }{ {(2 \times {10}^{6}) }^{2} } \\ \\ g = \frac{6.67 \times {10}^{ (- 11 + 24)} }{2 \times 2 \times {10}^{12} } \\ \\ g = \frac{6.67 \times {10}^{13} }{4 \times {10}^{12} } \\ \\ g = \frac{6.67}{4} \times 10 \\ \\ g = \frac{66.7}{4} \\ \\ g = 16.65 \: \frac{m}{ {s}^{2} }$

ANSWER:

The acceleration due to gravity on the imaginary planet is 16.65 m/s².

ACCELERATION DUE TO GRAVITY:

• It is directly proportional to the mass of the planet and is independent of the mass of the object on which it is acting .

• It is inversely proportional to the square of the radius of the planet.

• Acceleration due to gravity always acts towards the centre of the planet.

• It also varies due to lattitude effect, rotational effect, height and depth at which it is measured.