Question

If a planet has 3 times the radius of the Earth, but has the same density as the Earth, what is the gravitational acceleration at the surface of the planet? (g = 9.8 m/s2 )

Answer 1

Answer:

The Gravitational Acceleration of the Big planet is G = 3g

where G is the gravitational acceleration of the Big Planet and g is the gravitational acceleration of the Earth

So, it becomes G = 3 x 9.8$ms^{-2}$

G = 29.4 $ms^{-2}$  

Explanation:

Let's start by calculating the Density of the Earth of mass m anad Radius r

We know that mass is density times the volume

ρ = $\frac{m}{V}$

m = ρV

we know the volume is V = $\frac{4}{3}$ $πr^{3}$ (please ignore the symbol of Pi)

m = ρ$\frac{4}{3}$ $πr^{3}$

Calculating the Density of Big Planet

→For Big Planet we know that radius is 3-times the radius of Earth, that is 3r

M = ρ$\frac{108}{3}$ $π(r)^{3}$    

So in conclusion, the mass of Earth and the mass of Big planet are related as M = 27m    

Now let's come to the Gravitational Force, we know that gravitational force is directly proportional to the mass of the body and inversely proportional to the radius.

→Suppose for Earth:  Mass = m, Radius = r  

For Big Planet:   Mass = M, Radius = R

$\frac{m}{r^{2} }$ :  $\frac{M}{R^{2} }$ = g : G

$\frac{Gm}{r^{2} }$ =  $\frac{gM}{R^{2} }$

G =$\frac{gMr^{2}}{mR^{2}}$

→ Putting the value of R = 3r and M = 27m

G = $\frac{g27mr^{2}}{m(3r)^{2}}$

By cutting the terms, we get

G = 3g

value of g= 9.8$ms^{-2}$  

G = 3 x $9.8ms^{-2}$  

G = 29.4 $ms^{-2}$