Question

What is the magnitude of the gravitational force between the earth and a 1kg object on its surface? (Massof the earth is 6×10 raise power 24kg and radius of the earth is 6×10raise power 6m

Answer 1

Given :-

• Mass of object on the surface of earth , m₁ = 1 kg
• Mass of Earth , m₂ = 6 × 10²⁴ kg
• Radius of earth , R = 6 × 10⁶ m

To find :-

• Magnitude of Gravitational force between the earth and 1 kg object on its surface ( F )

Knowledge required :-

• Formula for Universal Law of gravitation

$\orange{\bigstar}\boxed{\sf{F=G\dfrac{m_1\times m_2}{R^2}}}$

(where F is the Magnitude of gravitational force between two bodies of masses m₁ and m₂ , G is the gravitational constant and R is the distance between two bodies)

• Value of Universal gravitational constant

$\orange{\bigstar}\boxed{\sf{G=6.67 \times 10^{-11}\;\;\;\;kg^{-2}\;N\;m^2}}$

Solution :-

Since, object of mass 1 kg is on the surface of earth , therefore distance between the earth and object will be equal to the radius of earth .

Now,

Using Formula for universal Law of gravitation

$\implies\sf{F=G\dfrac{m_1\times m_2}{R^2}}\\\\\\\sf{putting\:values\:in\:RHS}\\\\\\\implies\sf{F=\dfrac{(6.67\times 10^{-11}) \times( 1)\times( 6 \times 10^{24})}{(6\times 10^{6})^2}}\\\\\\\implies\sf{F=\dfrac{6.67\times 10^{-11}\times 6\times 10^{24}}{6\times 10^{6}\times 6\times 10^{6}}}\\\\\\\implies\sf{F=\dfrac{6.67\times 10^{13}}{6\times 10^{12}}}\\\\\\\implies\sf{F=1.11 \times 10}\\\\\\\implies\underline{\underline{\boxed{\large{\red{\sf{F=11\;\;N}}}\;\;\;\sf{(approx.)}}}}$

Therefore,

There will be a Gravitational force of magnitude 11 Newton between the earth and 1 Kg object.

Answer 2

Explanation:

F=G

R

2

m

1

×m

2

puttingvaluesinRHS

⟹F=

(6×10

6

)

2

(6.67×10

−11

)×(1)×(6×10

24

)

⟹F=

6×10

6

×6×10

6

6.67×10

−11

×6×10

24

⟹F=

6×10

12

6.67×10

13

⟹F=1.11×10

⟹

F=11N(approx.)