Question

Question :-

The density of the planet Mars is 3.93 g cm^-3 and its radius is 3,390 km . it is assumed that density distribution is uniform throughout the Mars , then calculate the value of acceleration due to gravity at the surface of Mars.

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Answer 1

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Here is the solution

___________________▶▶▶

Given:-

Density of Planet Mars = 3.93 g cm^-3 = 3.93 × 10³ kg m^-3

Radius = 3,390 km = 3.39 × 10^6 m

Find:-

acceleration due to gravity of mars (g) = ?

mass of mars (m) = ?

Solution▶▶

=========

As we already know that,

density = mass / Volume

we can also write it as

mass = density × volume

mass = ( 3.93 × 10³ kg m^-3)[(4/3)πR³]

As we know that, volume of Sphere = (4/3)πR³

then put the values..

=(3.93 × 10³ kg m^-3) × [(4/3) × 3.14 × (3.39 × 10^6m)³]

After solving this we get,

M = 6.41 × 10^23 kg

where,

mass of mars (M)

Using,

g = GM/R²

Therefore ,

acceleration due to gravity at the surface of Mars is

g =

(6.67 × 10^-11 N m² kg^-2)(6.41×10^23kg) / (3.39 × 10^6 m)²

after solving this we get,

= 3.71 m / s²

Acceleration due to gravity of Mars is

= 3.71 m/s²

Hope it helps you out▶▶▶

Answer 2

Here is the solution

___________________▶▶▶

Given:-

Density of Planet Mars = 3.93 g cm^-3 = 3.93 × 10³ kg m^-3

Radius = 3,390 km = 3.39 × 10^6 m

Find:-

acceleration due to gravity of mars (g) = ?

mass of mars (m) = ?

Solution▶▶

=========

As we already know that,

density = mass / Volume

we can also write it as

mass = density × volume

mass = ( 3.93 × 10³ kg m^-3)[(4/3)πR³]

As we know that, volume of Sphere = (4/3)πR³

then put the values..

=(3.93 × 10³ kg m^-3) × [(4/3) × 3.14 × (3.39 × 10^6m)³]

After solving this we get,

M = 6.41 × 10^23 kg

where,

mass of mars (M)

Using,

g = GM/R²

Therefore ,

acceleration due to gravity at the surface of Mars is

g =

(6.67 × 10^-11 N m² kg^-2)(6.41×10^23kg) / (3.39 × 10^6 m)²

after solving this we get,

= 3.71 m / s²

Acceleration due to gravity of Mars is

= 3.71 m/s²