Centripetal acceleration and universal gravitation problems

Question

1.A 0.039 kg rubber stopper is attached to a 1.0 m length of string. The stopper is swung in a horizontal circle with a velocity of 4 m/s. What is the centripetal force on the stopper?

2.Two asteroids have their centers 800 m apart. One has a mass of 31,000,000 kg. The other has a mass of 87,000,000 kg. What is the gravitational force between them?

3.How far apart are two satellites of equal mass, 1,420 kg, if the gravitational force between them is 5 x 10^-7 N?

4.Tom has a mass of 89 kg. He and Sally are standing 24 m apart. What is Sally’s mass if the force of gravitational attraction between them is 5 x 10^-10 N?

5.The radius of a planet is 8.33 x 10⁶ m. The mass of the planet is 2.67 x 10²³ kg. Find the acceleration due to gravity on the surface of the planet?

Answer 1

1. F = ma but the centripetal acceleration = v²/r so F = (mv²)/r
2. F = Gm₁m₂/d² where G = 6.674 x 10^-11 m³/kg/s².
3. Using F = Gm₁m₂/d² and solving for d and equating m₁ = m₂ = m, we get d² = Gm²/F so d = √(Gm²/F) or d = m√(G/F)
4. Using F = Gm₁m₂/d² and replacing m₁ with m_T (mass of Tom) and m₂ with m_S (mass of Sally) and solving for m_S, we get: F = Gm_Tm_S/d² or (Fd²)/(Gm_T) = m_S
5. Let g_P be the gravitational acceleration of the planet. Using F = Gm₁m₂/d², we can also say that F = ma, or in this case F = mg_P. We can replace m₁ with M_P (mass of the planet) and m₂ can be replaced with just m which represents the mass of some random object on the surface of the planet. So, GM_Pm/d² = mg_P. Since the mass of a planet acts as if all the mass was concentrated at the center, the distance from M_P to m is the radius of the planet (r), so GM_Pm/r² = mg_P. Dividing both sides by m, the m's just cancel out making the equation: GM_P/r² = g_P