Question

2. Consider a planet in some solar system which has a mass double the mass of the earth and
density equal to the average density of the earth. An object weighing Won the earth will
weigh
(a) W
(b) 2W
(c) W/2
(d) 21/3W at the planet
aglar
acer​

Answer 1

The object weighing W on Earth will weigh (d)$2^\frac{1}{3}\ W$ at the planet.

Explanation:

Formula:

• $Density = \frac{Mass}{Volume}$
• Volume of sphere = $\frac{4}{3} \pi R^3$
• Gravitational acceleration = $g = \frac{GM}{R^2}$  
• Weight $W = mg$  

Let

$m$ = Mass of the object

$M_e$ = Mass of Earth

$M_p$ = Mass of the planet

$D_e$ = Density of Earth

$D_p$ = Density of the planet

$V_e$ = Volume of Earth

$V_p$ = Volume of the planet

$R_e$ = Radius of Earth

$R_p$ = Radius of the planet

$g_e$ = Gravitational acceleration of Earth

$g_p$ = Gravitational acceleration of the planet

$W_e$ = Weight of object on Earth

$W_p$ = Weight of object on the planet

$G$ = Gravitational constant.

Given:

$M_p = 2M_e$ .......................(1)

$D_e = D_p$ ..........................(2)

$W_e = W\\\\W_p = ?$

Volume of Earth   :   $V_e = \frac{4}{3}\pi {(R_e)}^3$

Volume of planet   :   $V_p = \frac{4}{3}\pi {(R_p)}^3$

Density of Earth   :   $D_e=\frac{M_e}{V_e} =\frac{M_e}{\frac{4}{3}\pi {(R_e)}^3}$ .........................(3)

Density of planet   :   $D_p=\frac{M_p}{V_p} =\frac{M_p}{\frac{4}{3}\pi {(R_p)}^3}$ ........................(4)

From (2), $D_e = D_p$ , Equation (3) = Equation (4)  

$\frac{M_e}{\frac{4}{3}\pi {(R_e)}^3} = \frac{M_p}{\frac{4}{3}\pi {(R_p)}^3}\\\\\frac{M_p}{M_e} = \frac{\frac{4}{3}\pi {(R_p)}^3}{\frac{4}{3}\pi {(R_e)}^3}\\\\\frac{M_p}{M_e} = (\frac{R_p}{R_e}) ^3 \\\\(\frac{R_p}{R_e})=(\frac{M_p}{M_e})^{\frac{1}{3} }$..................(5)

Acceleration due to gravity of Earth : $g_e = \frac{GM_e}{{R_e}^2}$ ............. (6)

Acceleration due to gravity of planet : $g_p = \frac{GM_p}{{R_p}^2}$ ............. (7)

Dividing (7) by (6), we get

$\frac{g_p}{g_e} =\frac{ \frac{GM_p}{{R_p}^2}}{\frac{GM_e}{{R_e}^2}} \\\\\frac{g_p}{g_e} = \frac{M_p}{M_e}\times {\frac{{R_e}^2} {{R_p}^2} \\\\= \frac{M_p}{M_e}\times {({\frac{{R_e}} {{R_p}})^2}$

$\frac{g_p}{g_e} = \frac{M_p}{M_e}\times (\frac{1} {\frac{{R_p}} {{R_e}}})^2}$.............(8)

From (5), equation (8) becomes

$\frac{g_p}{g_e} = \frac{M_p}{M_e}\times (\frac{M_e}{M_p})^\frac{2}{3}$

$\frac{g_p}{g_e} = (\frac{M_p}{M_e})^\frac{1}{3}$

From (1), we get

$\frac{g_p}{g_e} = (2)^\frac{1}{3}$ ...........................(9)

Since Weight = Mass of object × Gravitational acceleration

$W_p = m\times g_p\\W_e = m\times g_e$

Therefore

$\frac{W_p}{W_e} = \frac{g_p}{g_e} = (2)^\frac{1}{3}$ .............[From (9)]

$W_p = (2^\frac{1}{3}) W_e$

Since the object weighs W on earth

$W_p = (2^\frac{1}{3}) W$

Thus the object weighs $(2^\frac{1}{3}) W$ at  the planet.

Answer 2

Answer:

the correct answer is the option d

Explanation:

Since the density of planet is same as that of earth.

ρ

p

=ρ

e

⟹

3

4

πR

p

3

M

p

=

3

4

πR

e

3

M

e

⟹

R

e

R

p

=(

M

e

M

p

)

1/3

The value of gravitational acceleration=g=

R

2

GM

⟹

W

e

W

p

=

mg

e

mg

p

⟹

g

e

g

p

=

M

e

M

p

R

p

2

R

e

2

=

M

e

M

p

(

M

p

M

e

)

2/3

=(

M

e

M

p

)

1/3

=2

1/3

⟹W

p

=2

1/3

W