The planet Krypton has a mass of
6.8 × 1023 kg and radius of 2.8 × 106 m.
What is the acceleration of an object in free
fall near the surface of Krypton? The gravitational
constant is 6.6726 × 10−11 N · m2
/kg2
.
Answer in units of m/s
2
.
A satellite moves in a circular orbit around
the Earth at a speed of 5.3 km/s.
Determine the satellite’s altitude above
the surface of the Earth. Assume the
Earth is a homogeneous sphere of radius
6370 km and mass 5.98 × 1024 kg. The
value of the universal gravitational constant
is 6.67259 × 10−11 N · m2
/kg2
.
Answer in units of km
To calculate the acceleration of an object in free fall near the surface of a planet, you can use the formula for gravitational acceleration:
acceleration = (gravitational constant * planet mass) / (planet radius)²
Given:
- Gravitational constant (G) = 6.6726 × 10⁻¹¹ N·m²/kg²
- Planet mass (M) = 6.8 × 10²³ kg
- Planet radius (R) = 2.8 × 10⁶ m
Substituting the values into the formula, we get:
acceleration = (6.6726 × 10⁻¹¹ N·m²/kg² * 6.8 × 10²³ kg) / (2.8 × 10⁶ m)²
Simplifying the expression, we have:
acceleration = 19.45 m/s²
So, the acceleration of an object in free fall near the surface of Krypton is 19.45 m/s².
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To calculate the satellite's altitude above the surface of the Earth, you can use the formula for the gravitational force acting on the satellite:
gravitational force = (gravitational constant * satellite mass * Earth mass) / (distance from satellite to Earth center)²
The gravitational force is balanced by the centripetal force required to keep the satellite in a circular orbit:
gravitational force = (satellite mass * satellite velocity²) / (distance from satellite to Earth center)
Given:
- Gravitational constant (G) = 6.67259 × 10⁻¹¹ N·m²/kg²
- Earth mass (M) = 5.98 × 10²⁴ kg
- Earth radius (R) = 6370 km = 6.37 × 10⁶ m
- Satellite velocity (v) = 5.3 km/s = 5.3 × 10³ m/s
To find the distance from the satellite to the Earth's center, we can use the equation:
distance from satellite to Earth center = Earth radius + satellite altitude
Substituting the values into the formula, we get:
(gravitational constant * satellite mass * Earth mass) / (distance from satellite to Earth center)² = (satellite mass * satellite velocity²) / (distance from satellite to Earth center)
Simplifying the expression, we have:
gravitational constant * satellite mass * Earth mass = satellite mass * satellite velocity² * (distance from satellite to Earth center)²
solving for (distance from satellite to Earth center)², we get:
(distance from satellite to Earth center)² = (gravitational constant * Earth mass) / satellite velocity²
Substituting the values into the equation, we have:
(distance from satellite to Earth center)² = (6.67259 × 10⁻¹¹ N·m²/kg² * 5.98 × 10²⁴ kg) / (5.3 × 10³ m/s)²
Calculating, we find:
(distance from satellite to Earth center)² ≈ 8.947 × 10¹² m²
Taking the square root of both sides, we get:
distance from satellite to Earth center ≈ 2.993 × 10⁶ m
Since the distance from the satellite to the Earth's center is equal to the Earth's radius plus the satellite's altitude, we have:
Earth radius + satellite altitude = 2.993 × 10⁶ m
Substituting the known value for the Earth radius, we can solve for the satellite altitude:
6.37 × 10⁶ m + satellite altitude = 2.993 × 10⁶ m
Simplifying, we get:
satellite altitude ≈ -3.377 × 10⁶ m
However, a negative value for altitude doesn't make sense in this context, so we take the absolute value:
satellite altitude ≈ 3.377 × 10⁶ m
Converting it to kilometers, we have:
satellite altitude ≈ 3377 km
Therefore, the satellite's altitude above the surface of the Earth is approximately 3377 km.
1. a pure application for Newtons gravational equation. Use it.
2. Fg=Fcentripetal
GMe*m/(re+h)^2 = m*v^2/r=m*(5300m/s)^2/(re+h)
solve for h
re+h=G*Me/25e6)^2
checkthat, solve forh