The gravitational field strength at the surface of a planet is 9.8 N/kg. What is the gravitational potential at the surface if the radius of the planet is 6.4 x 10^6 m?
Practice Questions
1 question
Q1
The gravitational field strength at the surface of a planet is 9.8 N/kg. What is the gravitational potential at the surface if the radius of the planet is 6.4 x 10^6 m?
-62.72 x 10^6 J/kg
-9.8 J/kg
-19.6 x 10^6 J/kg
-39.2 x 10^6 J/kg
V = -g * r = -9.8 N/kg * 6.4 x 10^6 m = -62.72 x 10^6 J/kg.
Questions & Step-by-step Solutions
1 item
Q
Q: The gravitational field strength at the surface of a planet is 9.8 N/kg. What is the gravitational potential at the surface if the radius of the planet is 6.4 x 10^6 m?
Solution: V = -g * r = -9.8 N/kg * 6.4 x 10^6 m = -62.72 x 10^6 J/kg.
Steps: 6
Step 1: Understand that gravitational field strength (g) is given as 9.8 N/kg.
Step 2: Identify the radius (r) of the planet, which is 6.4 x 10^6 m.
Step 3: Use the formula for gravitational potential (V), which is V = -g * r.
Step 4: Substitute the values into the formula: V = -9.8 N/kg * 6.4 x 10^6 m.
Step 5: Calculate the multiplication: -9.8 * 6.4 = -62.72.
Step 6: Add the unit to the result: -62.72 x 10^6 J/kg.
