Question

The mass of the earth is 6 × 10²⁴ kg and that of the moon is 7.4 × 10²² kg. If the distance between the earth and the moon is 3.84 × 10⁵ km, calculate the force exerted by the earth on the moon.
( Take G = 6.7 × 10 ‐¹¹ Nm² kg -² )
​

Answer 1

• The mass of the earth, M = 6 × 10²⁴ kg.

• The mass of the moon, m = 7.4 × 10²² kg.

• The distance between the earth and the moon d = 3.84 × 10⁵ km.

$= 3.84 \times 10 {}^{5} \times 1000m = 3.84 \times 10 {}^{8} m$

According to the universal law of gravitational, the force exerted by the earth on the moon is given by

$F = G \: \frac{M \times m}{d {}^{2} }$

$\frac{(6.7 \times 10 {}^{ - 11} Nm {}^{2}kg {}^{ - 2} \times \: (6 \times 10 {}^{24} kg) \times (7.4 \times 10 {}^{22} kg}{3.84 \times 10 {}^{8} m {}^{2} }$

$= 2.02 \times 10 {}^{20} N$

$Thus, \: the \: force \: exerted \: by \: the \: earth \: \\ on \: the \: moon \: is \: 2.02 × 10 {}^{20} .</p><p>$

Answer 2

Answer:

The mass of the earth, M = 6 × 10²⁴ kg.

The mass of the moon, m = 7.4 × 10²² kg.

The distance between the earth and the moon d = 3.84 × 10⁵ km.

= 3.84 \times 10 {}^{5} \times 1000m = 3.84 \times 10 {}^{8} m=3.84×10

5

×1000m=3.84×10

8

m

According to the universal law of gravitational, the force exerted by the earth on the moon is given by

F = G \: \frac{M \times m}{d {}^{2} } F=G

d

2

M×m

\frac{(6.7 \times 10 {}^{ - 11} Nm {}^{2}kg {}^{ - 2} \times \: (6 \times 10 {}^{24} kg) \times (7.4 \times 10 {}^{22} kg}{3.84 \times 10 {}^{8} m {}^{2} }

3.84×10

8

m

2

(6.7×10

−11

Nm

2

kg

−2

×(6×10

24

kg)×(7.4×10

22

kg

= 2.02 \times 10 {}^{20} N=2.02×10

20

N

\begin{gathered}Thus, \: the \: force \: exerted \: by \: the \: earth \: \\ on \: the \: moon \: is \: 2.02 × 10 {}^{20} . < /p > < p > \end{gathered}

Thus,theforceexertedbytheearth

onthemoonis2.02×10

20

.</p><p>