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 )

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

G = 29.4  

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 =

m = ρV

we know the volume is V = (please ignore the symbol of Pi)

m = ρ

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    

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

:   = g : G

=  

G =

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

G =

By cutting the terms, we get

G = 3g

value of g= 9.8  

G = 3 x  

G = 29.4  

dicrrdirrkdirckrrdrrk ; jhgfdsgrthyjkul; i'lkjhgfdskhjl; jhgdfs

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete