Question

a body weight 1 kg on the surface of the moon is mass of the moon is 7.4×10 22 kg and radius of moon is 1740 km calculate the force acting between the body and the moon. acceleration product in the body. acceleration product in moon​

Answer 1

Explanation:

f=G (m1m2/r^2)

= 6.67×10^11 (1×7.4×10^22/1740)

=1.63×10^6N

Answer 2

Given :-

Mass of the body = 1 kg

Mass of the moon = 7.4 × 10²² kg

Radius of the moon = 1740 km

To Find :-

The force acting between the body and the moon.

The acceleration produced in the body.

The acceleration produced in the moon.

Analysis :-

Firstly in order to find the force acting between the body and the moon substitute the values given in the question using the universal law of gravitation.

Then using the formula of force find the acceleration produced in moon and body accordingly.

Solution :-

We know that,

• r = Distance
• f = Force
• m = Mass
• g = Gravity

Using the formula,

$\underline{\boxed{\sf Universal \ law \ of \ gravitation=\dfrac{GMm}{R^2} }}$

Given that,

Gravity (g) = 6.67 × 10⁻¹¹ N m²/kg²

Mass₁ (M) = 7.4 × 10²² kg

Mass₂ (m) = 1 kg

Radius (r) = 1740 km = 1740 × 1000

Substituting their values,

$\sf f=\dfrac{ 6.67 \times 10^{-11} \times 7.4 \times 10^{22} }{(1740 \times 1000)^2}$

$\sf f = \dfrac{ 49.358 \times 10^{11} }{ 300276 \times 10^8 }$

$\sf f=1.63 \ N$

Therefore, the force acting on the body is 1.63 N.

Using the formula,

$\underline{\boxed{\sf Force=Mass \times Acceleration}}$

CASE I (acceleration in the body) :-

Given that,

Force (f) = 1.63 N

Mass (m) = 1 kg

Substituting their values,

$\sf 1.63=1 \times a$

$\sf a=\dfrac{1.63}{1} =1.63 \ m/s$

Therefore, the acceleration produced by the body is 1.63 m/s.

CASE II (acceleration in the moon) :-

Given that,

Force (f) = 1.63 N

Mass (m) = 7.4 x10²² kg

Substituting their values,

$\sf 1.63 = 7.4 \times 10^{22} \times a$

$\sf a=\dfrac{1.63}{ 7.4 \times 10^{22} } =2.2 \times 10^{-23} \ m/s$

Therefore, the acceleration produced by the body is 2.2 × 10⁻²³ m/s.