Question

Solve question number (k) and (l)

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Answer 1

there you go

i am sure that first one is correct but i am not sure about second one

hope it helps

please ask if theres any doubt

Answer 2

Answer:

(k) Distance between them is 7.8 × 10¹¹ m.

(l) The force of attraction between them is 7.37 × 10⁸ N.

Explanation:

w.k.t., Force of attraction or Gravitational force between two masses m₁ and m₂ is $\tt{ F = \dfrac{G\:m_{1}\:m_{2}}{r^{2}} }$, where r is the distance between the two masses and G is Universal Gravitational Constant whose value is 6.67 × 10⁻¹¹ Nm²/kg².

(k) Given: Mass of Jupiter = m₁ = 1.9 × 10²⁷ kg ; Mass of sun = m₂ = 2 × 10³⁰ kg and Gravitational force between them = F = 4.166 × 10²³ N.

Let r be the distance between them.

$\tt{ \implies 4.166\times10^{23} = \dfrac{6.67\times10^{-11} \times 1.9\times10^{27} \times 2\times10^{30}}{r^{2}} }$

$\tt{ \implies 4.166\times10^{23} = \dfrac{25.346\times10^{46}}{r^{2}} }$

$\tt{ \implies r^{2} = \dfrac{25.346\times10^{46}}{4.166\times10^{23}} }$

$\tt{ \implies r^{2} = 6.084\times10^{23} }$

$\tt{ \implies r = 7.8\times10^{11} \ m }$

⇒ Distance between Jupiter and Sun is 7.8 × 10¹¹ m.

(l) Given: Mass of man = m₁ = 75 kg ; Mass of earth = m₂ = 6 × 10²⁴ kg and Distance between them = Radius of earth = 6380 km.

$\tt{ \implies F = \dfrac{6.67\times10^{-11} \times 75 \times 6\times10^{24}}{(6380)^{2}} }$

$\tt{ \implies F = \dfrac{3001.5\times10^{13}}{4.07\times10^{7}} }$

$\tt{ \implies F = 737.469\times10^{6} }$

$\tt{ \implies F = 7.37\times10^{8} \ N }$

⇒ Force of attraction between man and earth is 7.37 × 10⁸ N.