Question

the magnitude of gravitational field at a distance 3/2a from the Centre..

a)root3GM/a square

b) 4GM/9a square

c) GM/3a square

d) root3GM/a square

Answer 1

Question :

The magnitude of a gravitational field at a distance 3/2 a from the center is .....

a. √3 GM / a²

b. 4GM / 9a²

c. GM / 3a²

d. √(3GM) / a²

Solution :

Gravitational field/force at a distance of r from the center is given by ,

$\bigstar\ \; \pink{\textsf{\textbf{F _\text g =\ \dfrac{\text{GM}}{\text{r ^\text 2 }} }}}$

where ,

• F₉ denotes Gravitational field
• G denotes gravitational constant
• r denotes radius

Gravitational constant = G

Mass of the object = M

Distance from center (r) =  $\sf \dfrac{3}{2}a$

$:\implies \sf F_g=\dfrac{GM}{r^2}$

$:\implies \sf F_g=\dfrac{GM}{\big(\frac{3}{2}a\big)^2}$

$:\implies \sf F_g=\dfrac{GM}{\frac{9}{4}a^2}$

$:\implies \green{\textsf{\textbf{F _\text{g} = \dfrac{\text{4GM}}{\text{9a}^\text{2}} }}}\ \; \bigstar$

Option b

Answer 2

Answer:

• Option (B) 4GM/9a²

Explanation:

Given that,

• Distance from center (r) = 3/2 a

As we know that,

$\purple \bigstar \: \: \bold \orange{ g = \frac{GM}{r^2}}$

Where,

• G = Gravitational constant
• M = Mass of the object
• r = Distance from center.

[ Putting values ]

$\leadsto \: \bold{g = \frac{GM}{( \frac{3}{2}a) {}^{2} } }$

$\leadsto \: \bold{g = \frac{GM}{ \frac{9}{4}a {}^{2} } }$

$\leadsto \: \bold \green{g = \frac{4GM}{9a {}^{2} } } \: \: \red \bigstar$

For information:

• F = GMm/d²
• F = ma