Question

A meteoroid is moving towards a planet. It has mass m = 0.86x10 kg and speed v1 = 4.7x10' m/s at distance R1 = 2.3x10 m from the center of the planet. The radius of the planet is R = 0.74x107m. The mass of the planet is M = 7.6x1025 kg. There is no air around the planet.

(a) Enter an expression for the gravitational potential energy PEof the meteoroid at R, in terms of defined quantities and the gravitational constant G.
Part (b) Calculate the value of PE1, in joules. HA 14%
Part (c) Enter an expression for the total energy E, of the meteoroid at R, in terms of defined quantities. A 14%
Part (d) Calculate the value of E1, in joules. A 14%
Part (e) Enter an expression for the total energy E of the meteoroid at R, the surface of the planet, in terms of defined quantities and v, the meteoroid's speed when it reaches the planet's surface.

Answer 1

Answer:

(a) PE = – G×M×m/R

(b) PE1 = –1.89×10¹⁷J

(c) E = 1/2mv² + (– G×M×m/R)

(d) E1 = 9.50×10²³J

(e) E = 1/2mv² + (– G×M×m/R)

Explanation:

Given

m = mass of meteoroid = 0.86×10⁹kg

v1 = speed of meteoroid at R1 = 4.7×10⁷m/s

R1 = distance of meteoroid from the center of the planet = 2.3×10⁷m

M = mass of the planet = 7.6×10²⁵kg

R = radius of planet = 0.74×10⁷m

(a) PE = – G×M×m/R

The above equation has a negative sign because the force of gravity between the planet and the meteoroid is directed inward towards the planet and the potential energy decreases in the direction of the increasing force. So the closer the meteoroid gets to the planet the lesser becomes it's potential energy.

(b) We are required to calculate the Potential energy at R1, PE1

PE1 = – G×M×m/R1 =

– (6.67×10-¹¹ × 7.6×10²⁵ × 0.86×10⁹)/(2.3×10⁷)

= –1.89×10¹⁷J

(c) E = total mechanical energy which is a sum of the potential and kinetic energy at any given position

E = KE + PE

E = 1/2mv² + – G×M×m/R

(d) E1 is the total mechanical energy at R1

E1 = 1/2mv1² + (– G×M×m/R1)

E1 = 1/2×0.86×10⁹×(4.7×10⁷)² +

(–6.67×10-¹¹ × 7.6×10²⁵ × 0.86×10⁹)/(2.3×10⁷)

E1 = 9.50×10²³J

(e) E = 1/2mv² + (– G×M×m/R)

Answer 2

(a) PE = $\frac{-G*M*m}{R}$

(b) PE₁ = –1.89×10¹⁷J

(c) E =

(d) E₁ = 9.50×10²³J

(e) E = $\frac{1}{2} mv^2 + (\frac{-G*M*m}{R} )$

Firstly, write the given values:

m = mass of meteoroid = 0.86×10⁹kg

v₁ = speed of meteoroid at R₁ = 4.7×10⁷m/s

R₁ = distance of meteoroid from the center of the planet = 2.3×10⁷m

M = mass of the planet = 7.6×10²⁵kg

R = radius of planet = 0.74×10⁷m

To find:

(a) Potential energy is given by the following expression:

PE = $\frac{-G*M*m}{R}$

• A negative sign is due to the force of gravity between the planet and the meteoroid is directed inward towards the planet.
• The closer the meteoroid gets to the planet the lesser becomes it's potential energy.

(b) To find the value of PE₁

PE₁ = $\frac{-G*M*m}{R_1}=$$\frac{6.67 * 1^{-11} * 7.6 * 10^{25} * 0.86 * 10^9 }{2.3 * 10^7} =-1.89 * 10^{17} J$

(c) An expression for the total energy E, of the meteoroid at R, in terms of defined quantities can be given as:

E = KE + PE

E = $\frac{1}{2} mv^2 + (\frac{-G*M*m}{R} )$

(d) E₁ is the total mechanical energy at R₁

E₁ = $\frac{1}{2} mv_1^2 + (\frac{-G*M*m}{R_1} )$

$E_1= \frac{1}{2} * 0.86* 10^9 * (4.7 * 10^7)^2 +\frac{(-6.67 * 10^{-11}* 7.6 * 10^{25} * 0.86* 10^9)}{2.3*10^7} =9.50 * 10^{23} J$

(e) Expression for the total energy E of the meteoroid at R, the surface of the planet in terms of defined quantities and v, the meteoroid's speed when it reaches the planet's surface will be:

E = $\frac{1}{2} mv^2 + (\frac{-G*M*m}{R} )$

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