Question

Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4×10^22kg and radius of 1.74×10^6 m (G=6.7×10^-11Nm^2Kg^2)

Answer 1

Hello Mate!

Hence, g = GM / R²

Here, value of G = 6.67 ^ -11

M = 7.4 × 10²² kg and R = 1.74 × 10^6 m

g = ( 6.7 × 10^-11 × 7.4 × 10²² )/( 1.74 × 10^6 )²

= ( 49.58 × 10¹¹ )/( 3.0276 × 10¹² )

= 16.4 × 1 / 10

=> 1.64 m/s²

Have great future ahead!

Answer 2

We have to find the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 10²² kg and radius 1.74 × 10⁶ m.

also given, universal gravitational constant, G = 6.7 × 10¯¹¹ Nm²/Kg²

According to Newton's law of Gravitational force,

The gravitational force between two bodies of masses m and M separated r distance to each other is given by,

$F=\frac{GmM}{r^2}$

also we know,

according to Newton's 2nd law of motion,

Force is the product of mass and acceleration.

let m experience F force due to gravity.

so, F = ma , where a is acceleration due to gravity.

from above equations we get,

$F=ma=\frac{GmM}{r^2}$

$\implies a=\frac{GM}{r^2}$

here, G = 6.7 × 10¯¹¹ Nm²/kg² , M = 7.4 × 10²² Kg

r = 1.74 × 10⁶ m

⇒a = (6.7 × 10¯¹¹ × 7.4 × 10²²)/(1.74 × 10⁶)²

= 16.36 × 10¯¹ m/s²

= 1.636 m/s²

Therefore the acceleration due to gravity is 1.636 m/s².