Question

Hey,

A spherical mass of 20 kg situated at the surface of the earth is attracted by another spherical mass of 150 kg with a force equal to the weight of 0.25 mg when the centres of the masses of 30 cm apart. Calculate the mass and mean density of the earth assuming the radius of the earth to be
$6 \times {10}^{8 \:} cm.$
​

Answer 1

Solution :-

Given :-

Mass 1 = 20 kg

Mass 2 = 150 kg

F = 0.25 × 10⁻⁶ kg × g

Distance = 0.3 m

So by using the equation :-

$F = \dfrac{G M_1M_2}{r^2}$

$F = \dfrac{ 6.67 \times 10^{-11} \times 20 \times 150 }{(0.3)^2}$

$F = \dfrac{ 6.67 \times 10^{-11} \times 3000 }{9 \times 10^{-2}}$

$F = \dfrac{ 6.67 \times 10^{-11} \times 1000 \times 10^{2}}{3 \times }$

$F = \dfrac{ 6.67 \times 10^{-6}}{3}$

Now as we know the value of F

$0.25 \times 10^{-6} (g) = \dfrac{ 6.67 \times 10^{-6}}{3}$

$g = \dfrac{ 4 \times 6.67 }{3}$

So now as we know.

$g = \dfrac{G M_{Earth}}{radius^2}$

$\dfrac{ 4 \times 6.67 }{3} = \dfrac{6.67 \times 10^{-11} M_{earth}}{(6 \times 10^{8})^2}$

$\dfrac{4}{3} = \dfrac{M_{earth} \times 10^{-11}}{36 \times 10^{16}}$

$4 = \dfrac{M_{earth} \times 10^{-27}}{12}$

$M_{earth} = 48 \times 10^{27}$

$M_{earth} = 4.8 \times 10^{28}$

Now Density :- Consider attached.

Answer 2

Explanation:

Given :-

Mass 1 = 20 kg

Mass 2 = 150 kg

F = 0.25 × 10⁻⁶ kg × g

Distance = 0.3 m

So by using the equation :-

F = \dfrac{G M_1M_2}{r^2}F=

r

2

GM

1

M

2

F = \dfrac{ 6.67 \times 10^{-11} \times 20 \times 150 }{(0.3)^2}F=

(0.3)

2

6.67×10

−11

×20×150

F = \dfrac{ 6.67 \times 10^{-11} \times 3000 }{9 \times 10^{-2}}F=

9×10

−2

6.67×10

−11

×3000

F = \dfrac{ 6.67 \times 10^{-11} \times 1000 \times 10^{2}}{3 \times }F=

3×

6.67×10

−11

×1000×10

2

F = \dfrac{ 6.67 \times 10^{-6}}{3}F=

3

6.67×10

−6

Now as we know the value of F

0.25 \times 10^{-6} (g) = \dfrac{ 6.67 \times 10^{-6}}{3}0.25×10

−6

(g)=

3

6.67×10

−6

g = \dfrac{ 4 \times 6.67 }{3}g=

3

4×6.67

So now as we know.

g = \dfrac{G M_{Earth}}{radius^2}g=

radius

2

GM

Earth

\dfrac{ 4 \times 6.67 }{3} = \dfrac{6.67 \times 10^{-11} M_{earth}}{(6 \times 10^{8})^2}

3

4×6.67

=

(6×10

8

)

2

6.67×10

−11

M

earth

\dfrac{4}{3} = \dfrac{M_{earth} \times 10^{-11}}{36 \times 10^{16}}

3

4

=

36×10

16

M

earth

×10

−11

4 = \dfrac{M_{earth} \times 10^{-27}}{12}4=

12

M

earth

×10

−27

M_{earth} = 48 \times 10^{27}M

earth

=48×10

27

M_{earth} = 4.8 \times 10^{28}M

earth

=4.8×10

28